Answer
$$\frac{4}{3}$$
Work Step by Step
\begin{aligned}
\int_{-1}^{1} \int_{0}^{\pi} x^{2} \sin y d y d x &=\left.\int_{-1}^{1} x^{2}(-\cos y)\right|_{y=0} ^{\pi} d x\\
&=\int_{-1}^{1} x^{2}(-\cos \pi+\cos 0) d x \\
&=\int_{-1}^{1} x^{2}(1+1) d x=\int_{-1}^{1} 2 x^{2} d x\\
&=\left.\frac{2}{3} x^{3}\right|_{-1} ^{1}=\frac{2}{3}\left(1^{3}-(-1)^{3}\right)\\
&=\frac{4}{3}
\end{aligned}
