Evaluating integral by changing to polar coordinates $\iint_D x^2y \;dA$ where $D$ is "the top half of the disk with center the origin and radius $5$."
I got:
$\iint_D (x^2y) dA$
$\int^\pi_0\int^5_0(rcos^2(\theta)r\sin(\theta))rdrd\theta$
$\int^\pi_0\int^5_0r^3(cos^2(\theta)\sin(\theta))drd\theta$
$\int^5_0 r^3(\cos^2(\theta)\sin(\theta))dr = \frac{625}{4}\cos^2(\theta)\sin(\theta)$
$\frac{625}{4}\int^{\pi}_0\cos^2(\theta)\sin(\theta)d\theta= \frac{625}{12}\cos(\theta)^3 |^\pi_0=\frac{-625}{12}-\frac{625}{12}=\frac{-625}{6}$
Using substitution with $u=\cos(\theta),$ and $ du=-\sin(\theta)d\theta$
But the book's answer says $\frac{-1250}{3}$.