# 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#?

Hello !

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

Explanations.

1) The area of a circle of radius

Here you have to calculate area of

2) The equation

The equation

Remark.

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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:

- Express the equations of the circles in polar coordinates.
- Determine the limits of integration for the angle (\theta).
- Set up the integral for the area using polar coordinates.
- 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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