Answer
$$\frac{8}{3}$$
Work Step by Step
$$\eqalign{
& y = 1 - {x^2} \cr
& y = 0 \cr
& 1 - {x^2} = 0 \to x = \pm 1 \cr
& {\text{From the graph we can define the limits of integration}} \cr
& 0 \leqslant y \leqslant 1 - {x^2},{\text{ }} - 1 \leqslant x \leqslant 1 \cr
& {\text{Therefore}} \cr
& \int_a^b {\int_{g\left( x \right)}^{h\left( x \right)} {f\left( {x,y} \right)} dy} dx \cr
& = \int_{ - 1}^1 {\int_0^{1 - {x^2}} 2 dy} dx \cr
& = \int_{ - 1}^1 {\left[ {\int_0^{1 - {x^2}} 2 dy} \right]} dx \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{1 - {x^2}} 2 dy = \left[ {2y} \right]_{y = 0}^{y = 1 - {x^2}} \cr
& = \left[ {2\left( {1 - {x^2}} \right) - 2\left( 0 \right)} \right] \cr
& = 2\left( {1 - {x^2}} \right) \cr
& {\text{Therefore,}} \cr
& \int_{ - 1}^1 {\left[ {\int_0^{1 - {x^2}} 2 dy} \right]} dx = \int_{ - 1}^1 {2\left( {1 - {x^2}} \right)} dx \cr
& = 2\int_{ - 1}^1 {\left( {1 - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& = 2\left[ {x - \frac{1}{3}{x^3}} \right]_{ - 1}^1 \cr
& = 2\left[ {1 - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - 2\left[ { - 1 - \frac{1}{3}{{\left( { - 1} \right)}^3}} \right] \cr
& = 2\left( {\frac{2}{3}} \right) - 2\left( { - \frac{2}{3}} \right) \cr
& = \frac{8}{3} \cr} $$
