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How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2<=x<=8#?

Answer 1

Scarlett Allison

graph{y = sqrt(8-x) [-1.07, 10.97, -1.25, 5.76]} Rotating about x-axis:

#dS = 2 pi y ds#

Arc length is:

#ds = sqrt(1 + x'^2) dy#

#y = sqrt (8-x)qquad x = 8- y^2 qquad x' = - 2y #

#x: 2 rarr 8 qquad harr qquad y: sqrt6 rarr 0 #

# S = 2 pi int_C y sqrt(1 + x'^2) dy#

# = 2 pi int_0^sqrt6 y sqrt(1 + 4y^2) dy#

# = 2 pi int_0^sqrt6 d/dy(1/12(1 + 4y^2)^(3/2)) dy#

# = pi/6 [(1 + 4y^2)^(3/2)]_0^sqrt6 #

# = pi/6( [(1 + 24)^(3/2)] - [(1 + 0)^(3/2)] )#

#=(62 pi)/3#

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Answer 2

Ezekiel Alexander

To find the exact area of the surface obtained by rotating the curve ( y = \sqrt{8 - x} ) about the x-axis over the interval ( 2 \leq x \leq 8 ), you can use the formula for the surface area of revolution, which is ( S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ). Substituting ( y = \sqrt{8 - x} ), you can find ( \frac{dy}{dx} ). Then, plug these values into the formula and evaluate the integral from ( x = 2 ) to ( x = 8 ) to find the surface area.

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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.

How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2&lt;=x&lt;=8#?

How do you find the exact area of the surface obtained rotating the curve about the #x#-axis of #y=sqrt(8-x)#, #2<=x<=8#?