Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^4 \left( {\mathop \smallint \limits_{y = x/2}^2 {x^2}y{\rm{d}}y} \right){\rm{d}}x = \frac{{256}}{{15}}$
Work Step by Step
Referring to Figure 24 (B), we can consider the shaded domain as a vertically simple region. The lower boundary of the region is a line $y = \frac{x}{2}$.
So, we obtain the domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 4,\frac{x}{2} \le y \le 2} \right\}$
Next, we compute the double integral of $f\left( {x,y} \right) = {x^2}y$ over ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}y{\rm{d}}A = \mathop \smallint \limits_{x = 0}^4 \left( {\mathop \smallint \limits_{y = x/2}^2 {x^2}y{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = 0}^4 {x^2}\left( {\frac{1}{2}{y^2}|_{x/2}^2} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^4 {x^2}\left( {4 - \frac{{{x^2}}}{4}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{x = 0}^4 \left( {4{x^2} - \frac{{{x^4}}}{4}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{4}{3}{x^3} - \frac{1}{{20}}{x^5}} \right)|_0^4$
$ = \frac{1}{2}\left( {\frac{{256}}{3} - \frac{{256}}{5}} \right) = \frac{{256}}{{15}}$
