Answer
$$2 \pi$$
Work Step by Step
The integration area is a quarter of a circle or a radius of 2. Using the shell method, we get the volume as
\[
2 \pi=\frac{\pi}{2}\left[\frac{x^{4}}{4}\right]_{0}^{2}=\int_{0}^{2} \frac{\pi}{2} x \cdot x^{2} d x
\]
Using a double integral,
\[
\begin{aligned}
\iint_{R} f(x, y) d y d z &=\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}}\left(y^{2}+x^{2}\right) d y d x \\
&=\int_{0}^{2}\left[x^{2} y+\frac{y^{3}}{3}\right]_{0}^{\sqrt{4-x^{2}}} d x \\
&=\int_{0}^{2} \frac{2}{3} \sqrt{4-x^{2}}\left(x^{2}+2\right) d x \\
&=\left[\frac{2}{3}\left(\frac{1}{4} x\left(x^{2}+2\right) \sqrt{4-x^{2}}+6 \sin ^{-1}\left(\frac{x}{2}\right)\right)\right]_{0}^{2} \\
&=2 \pi
\end{aligned}
\]
