Answer
$$\frac{\pi}{2}
$$
Work Step by Step
Using shell method, we get the volume as
\[
\int_{0}^{1} 2 \pi x\left(-x^{2}+1\right) d x=2 \pi\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}=\frac{\pi}{2}
\]
Using a double integral
\[
\begin{aligned}
\iint_{R} f(x, y) d y d z &=\int_{-1}^{1} \int_{-\sqrt{1-x^{2}}}^{\sqrt{-x^{2}+1}}\left(-x^{2}-y^{2}+1\right) d y d x \\
&=\int_{-1}^{1}\left(2 \sqrt{-x^{2}+1}-2 x^{2} \sqrt{-x^{2}+1}-\frac{2}{3}(\sqrt{1-x^{2}})^{3}\right) \\
&=\int_{-1}^{1} \frac{4}{3}\left(-x^{2}+1\right) \sqrt{-x^{2}+1} d x \\
&=\frac{\pi}{2}
\end{aligned}
\]
