Answer
The coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {\frac{4}{5},0} \right)$.
Work Step by Step
We have the region bounded by ${y^2} = x + 4$ and $x=4$.
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 $x=4$. We find the lower and upper boundaries by intersecting the curve and the line. Substituting $x=4$ in ${y^2} = x + 4$ gives
${y^2} = 8$
$y = \pm \sqrt 8 $
Thus, we consider ${\cal D}$ as a horizontally simple region described by
${\cal D} = \left\{ {\left( {x,y} \right)| - \sqrt 8 \le y \le \sqrt 8 ,{y^2} - 4 \le x \le 4} \right\}$
By definition, the centroid is given by
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$, ${\ \ \ }$ $y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A$,
where $A$ is the area of ${\cal D}$.
Evaluate the area:
$A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \mathop \smallint \limits_{x = {y^2} - 4}^4 {\rm{d}}x{\rm{d}}y$
$ = \mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( {x|_{{y^2} - 4}^4} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( {8 - {y^2}} \right){\rm{d}}x$
$ = \left( {8y - \frac{1}{3}{y^3}} \right)|_{ - \sqrt 8 }^{\sqrt 8 }$
$ = 8\sqrt 8 - \frac{8}{3}\sqrt 8 + 8\sqrt 8 - \frac{8}{3}\sqrt 8 = \frac{{64\sqrt 2 }}{3}$
Evaluate the average of $x$-coordinate:
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \frac{3}{{64\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \mathop \smallint \limits_{x = {y^2} - 4}^4 x{\rm{d}}x{\rm{d}}y$
$ = \frac{3}{{128\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( {{x^2}|_{{y^2} - 4}^4} \right){\rm{d}}y$
$ = \frac{3}{{128\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( {16 - {y^4} + 8{y^2} - 16} \right){\rm{d}}y$
$ = \frac{3}{{128\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( { - {y^4} + 8{y^2}} \right){\rm{d}}y$
$ = \frac{3}{{128\sqrt 2 }}\left( {\left( { - \frac{1}{5}{y^5} + \frac{8}{3}{y^3}} \right)|_{ - \sqrt 8 }^{\sqrt 8 }} \right)$
$ = \frac{3}{{128\sqrt 2 }}\left( { - \frac{{64}}{5}\sqrt 8 + \frac{{64}}{3}\sqrt 8 - \frac{{64}}{5}\sqrt 8 + \frac{{64}}{3}\sqrt 8 } \right) = \frac{4}{5}$
Evaluate the average of $y$-coordinate:
$\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \frac{3}{{64\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \mathop \smallint \limits_{x = {y^2} - 4}^4 y{\rm{d}}x{\rm{d}}y$
$ = \frac{3}{{64\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } y\left( {x|_{{y^2} - 4}^4} \right){\rm{d}}y$
$ = \frac{3}{{64\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } y\left( {8 - {y^2}} \right){\rm{d}}y$
$ = \frac{3}{{64\sqrt 2 }}\mathop \smallint \limits_{y = - \sqrt 8 }^{\sqrt 8 } \left( {8y - {y^3}} \right){\rm{d}}y$
$ = \frac{3}{{64\sqrt 2 }}\left( {\left( {4{y^2} - \frac{1}{4}{y^4}} \right)|_{ - \sqrt 8 }^{\sqrt 8 }} \right)$
$ = \frac{3}{{64\sqrt 2 }}\left( {32 - 16 - 32 + 16} \right) = 0$
So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {\frac{4}{5},0} \right)$.
