Answer
\[V=\frac{2}{5}\]
Work Step by Step
\[\begin{align}
& \text{Let the region }R=\left\{ \left( x,y \right):0\le y\le 1-{{x}^{2}},\text{ 0}\le x\le 1\text{ } \right\} \\
& \text{The volume is given by} \\
& V=\int_{0}^{1}{\int_{0}^{1-{{x}^{2}}}{\left[ \left( 2-y \right)-1 \right]}}dydx \\
& V=\int_{0}^{1}{\int_{0}^{1-{{x}^{2}}}{\left( 1-y \right)}}dydx \\
& \text{Integrating} \\
& V=\int_{0}^{1}{\left[ y-\frac{1}{2}{{y}^{2}} \right]}_{0}^{1-{{x}^{2}}}dx \\
& V=\int_{0}^{1}{\left[ y-\frac{1}{2}{{\left( 1-{{x}^{2}} \right)}^{2}} \right]}dx \\
& V=\int_{0}^{1}{\left( 1-{{x}^{2}}-\frac{1}{2}\left( 1-2{{x}^{2}}+{{x}^{4}} \right) \right)}dx \\
& V=\int_{0}^{1}{\left( 1-{{x}^{2}}-\frac{1}{2}+{{x}^{2}}-\frac{1}{2}{{x}^{4}} \right)}dx \\
& V=\int_{0}^{1}{\left( \frac{1}{2}-\frac{1}{2}{{x}^{4}} \right)}dx \\
& V=\left[ \frac{1}{2}x-\frac{{{x}^{5}}}{10} \right]_{0}^{1} \\
& V=\frac{1}{2}-\frac{1}{10} \\
& V=\frac{2}{5} \\
\end{align}\]
