Answer
\[V=\frac{4}{3}\]
Work Step by Step
\[\begin{align}
& \text{The volume is given by} \\
& V=\int_{0}^{1}{\int_{0}^{2}{\left( 1-{{x}^{2}} \right)}dydx} \\
& V=\int_{0}^{1}{\left[ \left( 1-{{x}^{2}} \right)y \right]_{0}^{2}dx} \\
& V=\int_{0}^{1}{2\left( 1-{{x}^{2}} \right)dx} \\
& V=\left[ 2x-\frac{2{{x}^{3}}}{3} \right]_{0}^{1} \\
& \text{Evaluate} \\
& V=\left[ 2\left( 1 \right)-\frac{2{{\left( 1 \right)}^{3}}}{3} \right]-\left[ 2\left( 0 \right)-\frac{2{{\left( 0 \right)}^{3}}}{3} \right] \\
& V=\frac{4}{3} \\
\end{align}\]
