Answer
\[A=\frac{32}{3}\]
Work Step by Step
\[\begin{align}
& \text{From the graph} \\
& R=\left\{ \left( x,y \right):{{x}^{2}}\le y\le 4,\text{ }-\text{2}\le x\le 2\text{ } \right\} \\
& \text{Then,} \\
& A=\int_{-2}^{2}{\int_{{{x}^{2}}}^{4}{dy}dx} \\
& \text{Integrate} \\
& A=\int_{-2}^{2}{\left[ y \right]_{{{x}^{2}}}^{4}dx} \\
& A=\int_{-2}^{2}{\left( 4-{{x}^{2}} \right)dx} \\
& A=\left[ 4x-\frac{1}{3}{{x}^{3}} \right]_{-2}^{2} \\
& A=\left[ 4\left( 2 \right)-\frac{1}{3}{{\left( 2 \right)}^{3}} \right]-\left[ 4\left( -2 \right)-\frac{1}{3}{{\left( -2 \right)}^{3}} \right] \\
& A=\frac{16}{3}+\frac{16}{3} \\
& A=\frac{32}{3} \\
\end{align}\]
