What went wrong? Calculate mass given the density function 
Calculate the mass:
  $$D = \{1 \leq x^2 + y^2 \leq 4 , y \leq 0\},\quad p(x,y) = y^2.$$

So I said:

$M = \iint_{D} {y^2 dxdy} = [\text{polar coordinates}] = \int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3sin^2\theta dr}}$.

But when I calculated that I got the answer $0$ which is wrong, it should be $\frac{15\pi}{8}$. Can someone please tell me what I did wrong?
 A: You have the integral
$$M=\int_\pi^{2\pi} d\theta\int_1^2 r^3 \sin^2\theta dr $$
this seems fine. One thing is sure this integral is not zero: Indeed your can write it as a product of two integrals
$$M=(\int_\pi^{2\pi} \sin^2\theta d\theta) (\int_1^2 r^3  dr)$$
and both those integrals give strictly positive numbers.
I would advise you to compute these two integrals separately and check you get stricly positive numbers.
A: You have the set-up correct, but you have incorrectly computed the integral
Let's work it out together.
$\int_{\pi}^{2\pi}d\theta {\int_{1}^{2} {r^3\sin^2\theta dr}}$
$\int_{\pi}^{2\pi} {\int_{1}^{2} {r^3\sin^2\theta drd\theta}}$
$\int_{\pi}^{2\pi} \sin^2\theta d\theta {\int_{1}^{2} {r^3dr}}$
$\int_{\pi}^{2\pi} \sin^2\theta d\theta  (\frac{2^4}{4} - \frac{1^4}{4})$
$\int_{\pi}^{2\pi} \sin^2\theta d\theta  (3\frac{3}{4})$
$\frac{1}{2}((2\pi - \sin(2\pi)\cos(2\pi) - \pi +\sin(\pi)\cos(\pi))  (3\frac{3}{4})$ 
note that the integral of $\sin^2(x)$ = $\frac{1}{2}(x - \sin(x)\cos(x))$
$\frac{1}{2}(\pi)(3\frac{3}{4}) = \frac{15\pi}{8}$
