Answer
\[\frac{32}{3}\]
Work Step by Step
\[\begin{align}
& \int_{-2}^{2}{\int_{0}^{4-{{y}^{2}}}{dx}dy} \\
& \text{Find the integral} \\
& =\int_{-2}^{2}{\left( 4-{{y}^{2}} \right)dy} \\
& =\left[ 4y-\frac{1}{3}{{y}^{3}} \right]_{-2}^{2} \\
& =\left[ 4\left( 2 \right)-\frac{1}{3}{{\left( 2 \right)}^{3}} \right]-\left[ 4\left( -2 \right)-\frac{1}{3}{{\left( -2 \right)}^{3}} \right] \\
& =\frac{16}{3}-\left( -\frac{16}{3} \right) \\
& \frac{32}{3} \\
& \text{Using the graph to switch the order of integration} \\
& \int_{-2}^{2}{\int_{0}^{4-{{y}^{2}}}{dx}dy}=\int_{0}^{4}{\int_{-\sqrt{4-x}}^{\sqrt{4-x}}{dy}dx} \\
& =\int_{0}^{4}{\left( 2\sqrt{4-x} \right)dx} \\
& =-2\left[ \frac{{{\left( 4-x \right)}^{3/2}}}{3/2} \right]_{0}^{4} \\
& =-\frac{4}{3}\left[ {{\left( 4-x \right)}^{3/2}} \right]_{0}^{4} \\
& =-\frac{4}{3}\left[ {{\left( 4-4 \right)}^{3/2}}-{{\left( 4-0 \right)}^{3/2}} \right] \\
& =-\frac{4}{3}\left[ 0-8 \right] \\
& =\frac{32}{3} \\
\end{align}\]
