Answer
\[A=\frac{304}{3}\]
Work Step by Step
\[\begin{align}
& \text{Let }y=\left| x \right|\text{ and }y=20-{{x}^{2}} \\
& y=\left| x \right|\to y=-x,\text{ }x<0,\text{ }y=x,\text{ }x>0, \\
& \text{From the graph, the area is given by:} \\
& A=\int_{-4}^{0}{\int_{-x}^{20-{{x}^{2}}}{dy}dx}+\int_{0}^{4}{\int_{x}^{20-{{x}^{2}}}{dy}dx} \\
& \text{Or by symmetry} \\
& A=2\int_{0}^{4}{\int_{x}^{20-{{x}^{2}}}{dy}dx} \\
& \text{Integrating} \\
& A=2\int_{0}^{4}{\left( 20-{{x}^{2}}-x \right)dx} \\
& A=2\left[ 20x-\frac{1}{3}{{x}^{3}}-\frac{1}{2}{{x}^{2}} \right]_{0}^{4} \\
& A=2\left[ 20\left( 4 \right)-\frac{1}{3}{{\left( 4 \right)}^{3}}-\frac{1}{2}{{\left( 4 \right)}^{2}} \right]-2\left[ 0 \right] \\
& A=2\left( \frac{152}{3} \right) \\
& A=\frac{304}{3} \\
\end{align}\]
