Answer
$$\frac{4}{3}$$
Work Step by Step
$$\eqalign{
& x = 1 - {y^2} \cr
& x = 0 \cr
& 1 - {y^2} = 0 \to y = \pm 1 \cr
& {\text{From the graph we can define the limits of integration}} \cr
& 0 \leqslant x \leqslant 1 - {y^2},{\text{ }} - 1 \leqslant y \leqslant 1 \cr
& {\text{Therefore}} \cr
& \int_a^b {\int_{g\left( y \right)}^{h\left( y \right)} {f\left( {x,y} \right)} dx} dy \cr
& = \int_{ - 1}^1 {\int_0^{1 - {y^2}} {\left( {1 + y} \right)} dx} dy \cr
& = \int_{ - 1}^1 {\left( {1 - y} \right)\left[ {\int_0^{1 - {y^2}} {dx} } \right]} dy \cr
& {\text{Evaluate inner integral}} \cr
& \int_0^{1 - {y^2}} {dx} = \left[ x \right]_{x = 0}^{x = 1 - {y^2}} \cr
& = 1 - {y^2} \cr
& {\text{Therefore,}} \cr
& \int_{ - 1}^1 {\left( {1 - y} \right)\left[ {\int_0^{1 - {y^2}} {dx} } \right]} dy \cr
& = \int_{ - 1}^1 {\left( {1 - y} \right)\left( {1 - {y^2}} \right)} dy \cr
& {\text{Expanding}} \cr
& = \int_{ - 1}^1 {\left( {1 - {y^2} - y + {y^3}} \right)} dy \cr
& {\text{Integrating}} \cr
& = \left[ {y - \frac{1}{3}{y^3} - \frac{1}{2}{y^2} + \frac{1}{4}{y^4}} \right]_{ - 1}^1 \cr
& = \left[ {1 - \frac{1}{3} - \frac{1}{2} + \frac{1}{4}} \right] - \left[ {\left( { - 1} \right) - \frac{1}{3}{{\left( { - 1} \right)}^3} - \frac{1}{2}{{\left( { - 1} \right)}^2} + \frac{1}{4}{{\left( { - 1} \right)}^4}} \right] \cr
& = \left( {\frac{5}{{12}}} \right) - \left( { - \frac{{11}}{{12}}} \right) \cr
& = \frac{4}{3} \cr} $$
