Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = {x^2}}^{4 - {x^2}} y{\rm{d}}y} \right){\rm{d}}x = \frac{{20}}{3}$
Work Step by Step
The inequalities describe the domain ${\cal D}$ as a vertically simple region. Please see the figure attached.
Next, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = {x^2}}^{4 - {x^2}} y{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{2}{y^2}|_{{x^2}}^{4 - {x^2}}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{{\left( {4 - {x^2}} \right)}^2} - {x^4}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_0^1 \left( {16 - 8{x^2}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {16x - \frac{8}{3}{x^3}} \right)|_0^1$
$ = \frac{1}{2}\left( {16 - \frac{8}{3}} \right) = \frac{{20}}{3}$
