Evaluate an integral using polar $\int\limits_0^2 \int\limits_{-4\sqrt{4-x^2}}^{4\sqrt{4-x^2}}(x^2-y^2)\,dy\,dx$

Question

How do you evaluate the following integral using polar cordinates.

$$\int_0^2 \int_{-4\sqrt{4-x^2}}^{4\sqrt{4-x^2}}(x^2-y^2)\:\mathrm{d}y\:\mathrm{d}x$$

I converted it to polar coordinate making it a circle with radius of 2.

$$r^2=x^2+y^2$$

I did

$$\int_0^2 (-1)(r^2)(r)\:\mathrm{d}r$$

which equals $-4$. Then the next step

$$\int_\frac{3\pi}{2}^\frac{\pi}{2}-4\: \mathrm{d}\theta$$

Then I get $4\theta$. Plug in my values but I am not sugesting the correct answer $0$.

Answer 1

Hint:

The correct transformation is $$ x=2\cos\theta $$ and $$ y=2r\sin\theta $$ The limits of integral are $0\le\theta\le\dfrac{\pi}{2}$ and $-4\le r\le 4$. The Jacobian is $4\sin^2\theta$. Try it! It works! If this hint doesn't help you, just comment below. I'll answer it for you. ヾ(-^〇^-)ﾉ

Answer 2

Set $$x=r\cos\varphi,\: y=r\sin\varphi$$

$$x^2+\frac{y^2}{16}=4\to r=\frac{2}{\sqrt{\cos^2\varphi+\frac{\sin^2\varphi}{16}}}$$

$$0\leq x\leq 2\to -\frac{\pi}{2}\leq \varphi\leq \frac{\pi}{2}$$

Hence $$I=\iint f(r,\varphi)rdrd\varphi=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\frac{2}{\sqrt{\cos^2\varphi+\frac{\sin^2\varphi}{16}}}} r^3\cos2\varphi drd\varphi$$

$$= 4\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos2\varphi}{\left(\cos^2\varphi+\frac{\sin^2\varphi}{16}\right)^2 }d\varphi=-120\pi$$