Answer
$\frac{1250}{3}$
Work Step by Step
We begin with the integral in rectangular coordinates
$$V=\iint_{D}x^2y\,dA$$
Using the substitutions for polar coordinates and adding the bounds for region $D$, we can rewrite the integral as:
$$V=\int_{0}^{\pi}\bigg(\int_{0}^{5}(r\cos\theta)^2r\sin\theta\,r\,dr\bigg)d\theta$$
Cleaning this up a bit, we get:
$$V=\int_{0}^{\pi}\bigg(\int_{0}^{5}r^4\cos^2\theta\sin\theta\,dr\bigg)d\theta$$
Solving the inner integral, we get:
$$V=\int_{0}^{\pi}\bigg[\frac{r^5}{5}\,cos^2\theta\sin\theta\bigg]_{0}^{5}d\theta=625\int_{0}^{\pi}\cos^2\theta\,\sin\theta\,d\theta$$
Using the substitution $u=\cos\theta$ and changing the bounds accordingly, we can rewrite the integral as:
$$V=625\int_{1}^{-1}-u^2du=625\int_{-1}^{1}u^2du$$
Solving, we get:
$$V=625\bigg[\frac{u^3}{3}\bigg]_{-1}^{1}=\frac{1250}{3}$$
