Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}{\rm{d}}A = \frac{{128}}{3}$
Work Step by Step
We divide the rhombus $\cal R$ into 4 regions.
Using the linearity properties of the double integral, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_I^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{II}^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{III}^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{IV}^{} {y^2}{\rm{d}}A$,
where $I$, $II$, $III$ and $IV$ denote the first, the second, the third and the fourth quadrant, respectively. We evaluate $f\left( {x,y} \right) = {y^2}$ over each quadrant separately and then sum them up to obtain $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}{\rm{d}}A$.
1. First quadrant:
We consider the domain in the first quadrant as a horizontally simple region. The right boundary is the line $y = - 2\left( {x - 2} \right) = - 2x + 4$. So, $x = \frac{{4 - y}}{2}$.
The domain description is given by
$I = \left\{ {\left( {x,y} \right)|0 \le y \le 4,0 \le x \le \frac{{4 - y}}{2}} \right\}$
We evaluate the double integral over $I$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_I^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{y = 0}^4 {y^2}\mathop \smallint \limits_{x = 0}^{\left( {4 - y} \right)/2} {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^4 {y^2}\left( {x|_0^{\left( {4 - y} \right)/2}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{y = 0}^4 {y^2}\left( {\frac{{4 - y}}{2}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^4 \left( {4{y^2} - {y^3}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{4}{3}{y^3} - \frac{1}{4}{y^4}} \right)|_0^4$
$ = \frac{1}{2}\left( {\frac{{256}}{3} - \frac{{256}}{4}} \right) = \frac{{32}}{3}$
2. Second quadrant:
We consider the domain in the second quadrant as a horizontally simple region. The left boundary is the line $y = 2\left( {x + 2} \right) = 2x + 4$. So, $x = \frac{{y - 4}}{2}$.
The domain description is given by
$II = \left\{ {\left( {x,y} \right)|0 \le y \le 4,\frac{{y - 4}}{2} \le x \le 0} \right\}$
We evaluate the double integral over $II$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{II}^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{y = 0}^4 {y^2}\mathop \smallint \limits_{x = \left( {y - 4} \right)/2}^0 {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^4 {y^2}\left( {x|_{\left( {y - 4} \right)/2}^0} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{y = 0}^4 {y^2}\left( {\frac{{4 - y}}{2}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{y = 0}^4 \left( {4{y^2} - {y^3}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{4}{3}{y^3} - \frac{1}{4}{y^4}} \right)|_0^4$
$ = \frac{1}{2}\left( {\frac{{256}}{3} - \frac{{256}}{4}} \right) = \frac{{32}}{3}$
3. Third quadrant:
We consider the domain in the third quadrant as a horizontally simple region. The left boundary is the line $y = - 2\left( {x + 2} \right) = - 2x - 4$. So, $x = - \frac{{y + 4}}{2}$.
The domain description is given by
$III = \left\{ {\left( {x,y} \right)| - 4 \le y \le 0, - \frac{{y + 4}}{2} \le x \le 0} \right\}$
We evaluate the double integral over $III$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{III}^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{y = - 4}^0 {y^2}\mathop \smallint \limits_{x = - \left( {y + 4} \right)/2}^0 {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 4}^0 {y^2}\left( {x|_{ - \left( {y + 4} \right)/2}^0} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{y = - 4}^0 {y^2}\left( {\frac{{y + 4}}{2}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{y = - 4}^0 \left( {{y^3} + 4{y^2}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{1}{4}{y^4} + \frac{4}{3}{y^3}} \right)|_{ - 4}^0$
$ = \frac{1}{2}\left( { - \frac{{256}}{4} + \frac{{256}}{3}} \right) = \frac{{32}}{3}$
4. Fourth quadrant:
We consider the domain in the fourth quadrant as a horizontally simple region. The right boundary is the line $y = 2\left( {x - 2} \right) = 2x - 4$. So, $x = \frac{{y + 4}}{2}$.
The domain description is given by
$IV = \left\{ {\left( {x,y} \right)| - 4 \le y \le 0,0 \le x \le \frac{{y + 4}}{2}} \right\}$
We evaluate the double integral over $IV$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{IV}^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{y = - 4}^0 {y^2}\mathop \smallint \limits_{x = 0}^{\left( {y + 4} \right)/2} {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 4}^0 {y^2}\left( {x|_0^{\left( {y + 4} \right)/2}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{y = - 4}^0 {y^2}\left( {\frac{{y + 4}}{2}} \right){\rm{d}}x$
$ = \frac{1}{2}\mathop \smallint \limits_{y = - 4}^0 \left( {{y^3} + 4{y^2}} \right){\rm{d}}x$
$ = \frac{1}{2}\left( {\frac{1}{4}{y^4} + \frac{4}{3}{y^3}} \right)|_{ - 4}^0$
$ = \frac{1}{2}\left( { - \frac{{256}}{4} + \frac{{256}}{3}} \right) = \frac{{32}}{3}$
Finally,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} {y^2}{\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_I^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{II}^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{III}^{} {y^2}{\rm{d}}A + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{IV}^{} {y^2}{\rm{d}}A$
$ = \frac{{32}}{3} + \frac{{32}}{3} + \frac{{32}}{3} + \frac{{32}}{3}$
$ = \frac{{128}}{3}$
