How do you find the center of mass if the density at any point is inversely proportional to its distance from the origin of a lamina that occupies the region inside the circle #x^2 + y^2 = 10y# but outside the circle #x^2+y^2=25#?

Answer 1

The enter of mass is #(0, 2.5)#

Consider the first equation:

# x^2+y^2=10y #

We can put this into standard from by completing the square:

# x^2 + y^2 - 10y = 0 #
# :. x^2 + (y-5)^2-5^2 = 0 #
# :. x^2 + (y-5)^2 = 5^2 #

Which is a circle of radius #5# and centre #(0,5)#, And now the second equation:

# x^2+y^2=25 #

Which is a circle of radius #5# and centre #(0,0)#

We can plot these curves;

For a more complex problem we would need to use integration to find the Centre of Mass, but because of symmetry the density will be evenly distributed about the lines #x=0# and #y=2.5#, these being the lines of symmetry.

Hence the enter of mass is #(0, 2.5)#

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Answer 2

To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the densityTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rhoTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density functionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function overTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the laminaTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dATo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina'sTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's regionTo find the center of mass of the lamina, you can use the formula:

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( \bar{x} = \frac{\int x \rho(x,y) dA}{\intTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide byTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rhoTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the totalTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total massTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

GivenTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dATo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given thatTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA}To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the densityTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

(To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any pointTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \barTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point isTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inverselyTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y}To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportionalTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} =To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional toTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to itsTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \fracTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distanceTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance fromTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\intTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the originTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin,To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the densityTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density functionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function \To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dATo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(xTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, yTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)\To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rhoTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y))To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) isTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/rTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dATo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r\To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA}To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r),To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), whereTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

WhereTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (kTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where (To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) isTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is aTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constantTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant andTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (rTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) )To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r)To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) isTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) isTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the densityTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance fromTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

ForTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the originTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For thisTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin toTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problemTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem,To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the pointTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point \To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the densityTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density functionTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((xTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x,To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rhoTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, yTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)\To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

ToTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculateTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the totalTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) )To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total massTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) isTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass,To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inverselyTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrateTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportionalTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional toTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the densityTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density functionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distanceTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function overTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance fromTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the regionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the originTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region ofTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin,To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, soTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so (To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the laminaTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rhoTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

ToTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To setTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set upTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y)To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \fracTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals,To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, youTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first needTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need toTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits ofTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 +To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integrationTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration byTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determiningTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}}To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersectionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection pointsTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solveTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the twoTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circlesTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integralsTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

ThenTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, youTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need toTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (xTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set upTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x)To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up doubleTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) andTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integralsTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y)To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals overTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multipliedTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied byTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the regionTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region ofTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the densityTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density functionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function overTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina.To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the regionTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. TheTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region definedTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The regionTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined byTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region isTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by theseTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is insideTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside theTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

FinallyTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle (To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divideTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 +To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results byTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the totalTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total massTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 =To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass toTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to findTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10yTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinatesTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y \To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates ofTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y )To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of theTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) butTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of the centerTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) but outsideTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of the center ofTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) but outside the circleTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of the center of massTo find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) but outside the circle (To find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of the center of mass.To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) but outside the circle ( xTo find the center of mass of the lamina, you need to integrate the position vector ((x, y)) multiplied by the density function over the lamina's region and divide by the total mass.

Given that the density at any point is inversely proportional to its distance from the origin, the density function (\delta(x, y)) is (k/r), where (k) is a constant and (r) is the distance from the origin to the point ((x, y)).

To calculate the total mass, integrate the density function over the region of the lamina.

To set up the integrals, you first need to find the limits of integration by determining the intersection points of the two circles.

Then, integrate (x) and (y) multiplied by the density function over the region defined by these limits.

Finally, divide the results by the total mass to find the coordinates of the center of mass.To find the center of mass of the lamina, you can use the formula:

( \bar{x} = \frac{\int x \rho(x,y) dA}{\int \rho(x,y) dA} )

( \bar{y} = \frac{\int y \rho(x,y) dA}{\int \rho(x,y) dA} )

Where ( \rho(x,y) ) is the density function.

For this problem, the density function ( \rho(x,y) ) is inversely proportional to the distance from the origin, so ( \rho(x,y) = \frac{1}{\sqrt{x^2 + y^2}} ).

To solve the integrals, you need to set up double integrals over the region of the lamina. The region is inside the circle ( x^2 + y^2 = 10y ) but outside the circle ( x^2 + y^2 = 25 ). You'll need to convert to polar coordinates to evaluate the integrals.

Once you have the integrals set up, you can evaluate them to find the values of ( \bar{x} ) and ( \bar{y} ), giving you the coordinates of the center of mass.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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