# Thread: Polar coordinates, Double Integral, Circles

1. ## Polar coordinates, Double Integral, Circles

The question is attached.

For this question, this is what I plotted ( but I drew it uglier ): http://www.wolframalpha.com/input/?i=plot+x^2+-+2x+%2B+y^2+%3D+0+and+x^2+%2B+y^2+%3D+4

Here is the double integral I evaluated but is incorrect: http://www.wolframalpha.com/input/?i=integral+from+0+to+pi/2+(integral+from+2cos(theta)+to+2+(r^3+-+4r)+dr)+dtheta

Could someone tell me why it is incorrect? I suspect it's the integrand itself but I am not 100% sure. I tried 0 = f(x,y) = x^2 + y^2 - 4 = r^2 - 4 prior to multiplying it by the r in front of dr d(theta).

Any help would be greatly appreciated!

2. ## Re: Polar coordinates, Double Integral, Circles

Do you have to use polar coordinates from the get-go? I would probably write the integrals in cartesian coordinates, and then use a trig substitution to evaluate the integrals. In cartesian coordinates, I get

$\displaystyle A=\int_{0}^{2}\sqrt{4-x^{2}}\,dx-\int_{0}^{2}\sqrt{1-(x-1)^{2}}\,dx.$

For the first integral, draw a right triangle, with $\displaystyle \theta$ as one of the acute angles, $\displaystyle 2$ as the hypotenuse, $\displaystyle x$ as the opposite side, and $\displaystyle \sqrt{4-x^{2}}$ as the adjacent side.

For the second integral, draw a right triangle, with $\displaystyle \varphi$ as one of the acute angles, $\displaystyle 1$ as the hypotenuse, $\displaystyle x-1$ as the opposite side, and $\displaystyle \sqrt{1-(x-1)^{2}}$ as the adjacent side.

Then you turn the crank.

As for doing it in polar coordinates, I have several comments as to your approach.

0. You don't have to do double integrals to find area using polar coordinates. If you have a region $\displaystyle R$ defined by the curve $\displaystyle r(\theta)$ between the values of $\displaystyle a\le\theta\le b,$ then the integral defining that area is given by

$\displaystyle A=\frac{1}{2}\int_{a}^{b}[r(\theta)]^{2}\,d\theta.$

1. The $\displaystyle \frac{1}{2}\int_{0}^{\pi/2}4\,d\theta$ integral is positive, not negative. You shouldn't actually need to compute this integral, unless your teacher requires it.

2. The smaller curve is defined by $\displaystyle r=2\cos(\theta).$ So that's what you plug into your area formula above, and is what you subtract. You shouldn't need to compute this integral, either.

3. If you must use calculus to solve the problem, then you must. However, I would definitely use geometry and known results to check your answer.