Answer
$$A = \frac{{16}}{3}$$
Work Step by Step
$$\eqalign{
& x = - y \Rightarrow y = - x \cr
& x = 2 - {y^2} \Rightarrow y = \sqrt {2 - x} \cr
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{ - 2}^2 {\left( {\sqrt {2 - x} - \left( { - x} \right)} \right)} dx \cr
& A = \int_{ - 2}^2 {\left( {{{\left( {2 - x} \right)}^{1/2}} + x} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - \frac{2}{3}{{\left( {2 - x} \right)}^{3/2}} + \frac{1}{2}{x^2}} \right]_{ - 2}^2 \cr
& {\text{Evaluate }} \cr
& A = \left[ { - \frac{2}{3}{{\left( {2 - 2} \right)}^{3/2}} + \frac{1}{2}{{\left( 2 \right)}^2}} \right] - \left[ { - \frac{2}{3}{{\left( {2 + 2} \right)}^{3/2}} + \frac{1}{2}{{\left( { - 2} \right)}^2}} \right] \cr
& {\text{Simplify}} \cr
& A = 2 - \left[ { - \frac{{10}}{3}} \right] \cr
& A = \frac{{16}}{3} \cr} $$