Stuck solving $\iint \frac{x^2}{(x^2+y^2)}dA$ over the region $R$, $R$ lies between $x^2+y^2=4$ and $x^2+y^2=49$ by changing to polar coordinates.

Question

I plugged in for $x=r \cos\theta$ and $y=r \sin\theta$ to change to polar coordinates

I set my bounds as $2 \leqq r\leqq 7$ and $0\leqq \Theta \leqq 2\pi$.

I simplified my integral from

$$\displaystyle\iint\frac{r^2 \sin^2\theta}{(r^2 \sin^2(\theta)+r^2 \cos^2(\theta))}r dA$$ to $$\iint r \sin^2 (\theta) d\theta dr$$. I run into my issue with my first integral, because I get $\frac{-r}{3} \cos^3(\Theta$) and I have to plug in 2$\pi$ and $0$, and I end up with $\frac{-r}{3}+\frac{r}{3}=0$.

Answer 1

$\int_0^{2\pi} \cos^{2}\theta d\theta =\frac 1 2 \int_0^{2\pi} [1+ \cos (2\theta)] d\theta =\pi+0=\pi$. Now multiply by $r$ and integrate from $2$ to $7$.

Answer 2

You have a mistake in your integral. First of all, it should have been $x^2 = r^2 \cos^2\theta$.

Also $ \int \displaystyle \sin^2\theta \ d\theta \ne \cos^3\theta$.

Write $\sin^2\theta$ as $\cfrac{1 - \cos2\theta}{2} \ $ or $ \ \cos^2\theta \ $ as $ \ \cfrac{1 + \cos2\theta}{2}$

But there is an easier way to do this. Due to symmetry of the region,

$\displaystyle \iint \cfrac{x^2}{x^2+y^2} \ dA = \iint \cfrac{y^2}{x^2+y^2} \ dA = \cfrac{1}{2} \iint dA$

$\displaystyle = \cfrac{1}{2} \int_0^{2\pi} \int_2^7 r \ dr \ d\theta$