How do you use polar coordinates to evaluate the integral which gives the area that lies in the first quadrant between the circles #x^2+y^2=36# and #x^2-6x+y^2=0#?

Answer 1

Hello !

Do you really need integrals to calculate these aera ? Maybe I did not understand your problem, but the aera is easy :

#\mathcal{A} = \frac{1}{4}\pi\cdot 6^2 - \frac{1}{2}\pi\cdot 3^2 = \frac{9}{2}\pi#.

Explanations.

1) The area of a circle of radius #R# is #\pi R^2#.

Here you have to calculate area of #1/4# of circle of radius 6 (so, #\frac{1}{4}\pi\cdot 6^2#) and area of half circle of radius 3 (so, \frac{1}{2}\pi\cdot 3^2).

2) The equation #x^2+y^2=36# represents the circle of radius #\sqrt{36} = 6# and center #O(0,0)#.

The equation #x^2-6x+y^2=0# is too #(x-3)^2+ y^2=9#, therefore it represents the circle of radius #\sqrt{9}=3# and center #I(3,0)#.

Remark. #(x-a)^2 + (y-b)^2 = r^2# (with #r>0#) represents the circle of radius #r# and center #I(a,b)#.

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Answer 2

To evaluate the integral giving the area between the circles (x^2 + y^2 = 36) and (x^2 - 6x + y^2 = 0) in the first quadrant using polar coordinates, follow these steps:

  1. Express the equations of the circles in polar coordinates.
  2. Determine the limits of integration for the angle (\theta).
  3. Set up the integral for the area using polar coordinates.
  4. Integrate to find the area.

For the first circle (x^2 + y^2 = 36), in polar coordinates, (x = r \cos(\theta)) and (y = r \sin(\theta)). Substitute these into the equation to get (r = 6).

For the second circle (x^2 - 6x + y^2 = 0), complete the square to rewrite it as ((x - 3)^2 + y^2 = 9). This corresponds to a circle centered at ((3, 0)) with radius 3. In polar coordinates, it is (r = 6 \cos(\theta)).

The limits of integration for (\theta) are from 0 to (\pi/2) because we're only interested in the first quadrant.

The area can be expressed as (\int_{0}^{\pi/2} \frac{1}{2} \left((6 \cos(\theta))^2 - 6^2\right) , d\theta).

Solve this integral to find the 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.

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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