Answer
$\iint_Dx^2ydA=\frac{1250}{3}$
Work Step by Step
Given: $D$ is the top half of the disk with center the origin and radius $5$
The region $D$ is the set of points $\{(r,\theta)|0\leq r\leq 5, 0\leq \theta\leq \pi\}$.
Evaluate the integral:
$\iint_Dx^2ydA=\int_0^5\int_0^\pi (r\cos\theta)^2(r\sin\theta)rd\theta dr$
$=\int_0^5\int_0^\pi r^4\sin\theta\cos^2\theta d\theta dr$
$=\int_0^5 \frac{-r^4\cos^3\theta}{3}]_0^\pi dr$
$=\int_0^5(\frac{-r^4(-1)^3}{3}-\frac{-r^4\cdot 1^3}{3})dr$
$=\int_0^5\frac{2r^4}{3}dr$
$=\frac{2r^5}{15}]_0^5$
$=\frac{2\cdot 5^5}{15}-0$
$=\frac{1250}{3}$
