Answer
The center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( { - \frac{4}{3},0} \right)$.
Work Step by Step
We have ${\cal D}$, the region bounded by ${y^2} = x + 4$ and $x=0$; and the mass density $\delta \left( {x,y} \right) = \left| y \right|$.
From the figure attached we see that the region is bounded left by the curve ${y^2} = x + 4$ and bounded right by the line $x=0$. We find the lower and upper boundaries by intersecting the curve and the line. Substituting $x=0$ in ${y^2} = x + 4$ gives
$y = \pm 2$
Thus, we consider ${\cal D}$ as a horizontally simple region described by
${\cal D} = \left\{ {\left( {x,y} \right)| - 2 \le y \le 2,{y^2} - 4 \le x \le 0} \right\}$
Step 1. Evaluate the mass of ${\cal D}$:
$M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 \left| y \right|{\rm{d}}x{\rm{d}}y$
Since $\left| y \right| = \left\{ {\begin{array}{*{20}{c}}
y&{{\rm{for}{\ }}y \ge 0}\\
{ - y}&{{\rm{for}{\ }}y < 0}
\end{array}} \right.$,
the integral becomes
$M = - \mathop \smallint \limits_{y = - 2}^0 \mathop \smallint \limits_{x = {y^2} - 4}^0 y{\rm{d}}x{\rm{d}}y + \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 y{\rm{d}}x{\rm{d}}y$
$ = - \mathop \smallint \limits_{y = - 2}^0 y\left( {x|_{{y^2} - 4}^0} \right){\rm{d}}y + \mathop \smallint \limits_{y = 0}^2 y\left( {x|_{{y^2} - 4}^0} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^0 y\left( {{y^2} - 4} \right){\rm{d}}y - \mathop \smallint \limits_{y = 0}^2 y\left( {{y^2} - 4} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = - 2}^0 \left( {{y^3} - 4y} \right){\rm{d}}y - \mathop \smallint \limits_{y = 0}^2 \left( {{y^3} - 4y} \right){\rm{d}}y$
$ = \left( {\frac{1}{4}{y^4} - 2{y^2}} \right)|_{ - 2}^0 - \left( {\frac{1}{4}{y^4} - 2{y^2}} \right)|_0^2$
$ = - \left( {4 - 8} \right) - \left( {4 - 8} \right) = 8$
Step 2. Evaluate the $x$-coordinate of the center of mass:
${x_{CM}} = \frac{{{M_y}}}{M} = \frac{1}{8}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \frac{1}{8}\mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 x\left| y \right|{\rm{d}}x{\rm{d}}y$
$ = - \frac{1}{8}\mathop \smallint \limits_{y = - 2}^0 \mathop \smallint \limits_{x = {y^2} - 4}^0 xy{\rm{d}}x{\rm{d}}y + \frac{1}{8}\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 xy{\rm{d}}x{\rm{d}}y$
$ = - \frac{1}{{16}}\mathop \smallint \limits_{y = - 2}^0 \left( {{x^2}|_{{y^2} - 4}^0} \right)y{\rm{d}}y + \frac{1}{{16}}\mathop \smallint \limits_{y = 0}^2 \left( {{x^2}|_{{y^2} - 4}^0} \right)y{\rm{d}}y$
$ = \frac{1}{{16}}\mathop \smallint \limits_{y = - 2}^0 \left( {{y^4} - 8{y^2} + 16} \right)y{\rm{d}}y - \frac{1}{{16}}\mathop \smallint \limits_{y = 0}^2 \left( {{y^4} - 8{y^2} + 16} \right)y{\rm{d}}y$
$ = \frac{1}{{16}}\mathop \smallint \limits_{y = - 2}^0 \left( {{y^5} - 8{y^3} + 16y} \right){\rm{d}}y - \frac{1}{{16}}\mathop \smallint \limits_{y = 0}^2 \left( {{y^5} - 8{y^3} + 16y} \right){\rm{d}}y$
$ = \frac{1}{{16}}\left( {\frac{1}{6}{y^6} - 2{y^4} + 8{y^2}} \right)|_{ - 2}^0 - \frac{1}{{16}}\left( {\frac{1}{6}{y^6} - 2{y^4} + 8{y^2}} \right)|_0^2$
$ = - \frac{1}{{16}}\left( {\frac{{64}}{6} - 32 + 32} \right) - \frac{1}{{16}}\left( {\frac{{64}}{6} - 32 + 32} \right) = - \frac{4}{3}$
Step 3. Evaluate the $y$-coordinate of the center of mass:
${y_{CM}} = \frac{{{M_x}}}{M} = \frac{1}{8}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A = \frac{1}{8}\mathop \smallint \limits_{y = - 2}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 y\left| y \right|{\rm{d}}x{\rm{d}}y$
$ = - \frac{1}{8}\mathop \smallint \limits_{y = - 2}^0 \mathop \smallint \limits_{x = {y^2} - 4}^0 {y^2}{\rm{d}}x{\rm{d}}y + \frac{1}{8}\mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = {y^2} - 4}^0 {y^2}{\rm{d}}x{\rm{d}}y$
$ = - \frac{1}{8}\mathop \smallint \limits_{y = - 2}^0 {y^2}\left( {x|_{{y^2} - 4}^0} \right){\rm{d}}y + \frac{1}{8}\mathop \smallint \limits_{y = 0}^2 {y^2}\left( {x|_{{y^2} - 4}^0} \right){\rm{d}}y$
$ = \frac{1}{8}\mathop \smallint \limits_{y = - 2}^0 \left( {{y^4} - 4{y^2}} \right){\rm{d}}y - \frac{1}{8}\mathop \smallint \limits_{y = 0}^2 \left( {{y^4} - 4{y^2}} \right){\rm{d}}y$
$ = \frac{1}{8}\left( {\frac{1}{5}{y^5} - \frac{4}{3}{y^3}} \right)|_{ - 2}^0 - \frac{1}{8}\left( {\frac{1}{5}{y^5} - \frac{4}{3}{y^3}} \right)|_0^2$
$ = - \frac{1}{8}\left( { - \frac{{32}}{5} + \frac{{32}}{3}} \right) - \frac{1}{8}\left( {\frac{{32}}{5} - \frac{{32}}{3}} \right) = 0$
Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( { - \frac{4}{3},0} \right)$.
