Answer
$\pi$
Work Step by Step
Let $g(r)=r,\quad h(\theta)=\sin^{2}(\theta)$, and apply formula (5).
$\displaystyle \int_{0}^{2}\int_{0}^{\pi}g(r)h(\theta)d\theta dr=\int_{0}^{2}rdr\int_{0}^{\pi}\sin^{2}\theta d\theta\quad \left[\begin{array}{l}
\cos 2\theta=1-2\sin^{2}\theta,\\
\sin^{2}\theta=(1-\cos 2\theta)/2
\end{array}\right]$
$=\displaystyle \int_{0}^{2}rdr\cdot\int_{0}^{\pi}\frac{1}{2}(1-\cos 2\theta)d\theta$
$=\displaystyle \int_{0}^{2}rdr\cdot(\frac{1}{2}\int_{0}^{\pi}d\theta-\int_{0}^{\pi}\cos 2\theta d\theta)$
$=[\displaystyle \frac{1}{2}r^{2}]_{0}^{2}\cdot\frac{1}{2}\{[\theta]_{0}^{\pi}-\frac{1}{2}\sin 2\theta]_{0}^{\pi}\}$
$=(2-0)\cdot \displaystyle \frac{1}{2}[(\pi-0-(\frac{1}{2}\sin 2\pi-\frac{1}{2}\sin 0)]$
$=2\displaystyle \cdot\frac{1}{2}(\pi)$
$=\pi$
