Answer
The coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {0,\frac{2}{5}} \right)$.
Work Step by Step
We have the region bounded by $y = 1 - {x^2}$ and $y=0$.
From the figure attached we see that the region is bounded left by $x=-1$ and bounded right by $x=1$. Thus, the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {x,y} \right)| - 1 \le x \le 1,0 \le y \le 1 - {x^2}} \right\}$
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:
$A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} {\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {y|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = \mathop \smallint \limits_{x = - 1}^1 \left( {1 - {x^2}} \right){\rm{d}}x$
$ = \left( {x - \frac{1}{3}{x^3}} \right)|_{ - 1}^1 = \left( {1 - \frac{1}{3} + 1 - \frac{1}{3}} \right) = \frac{4}{3}$
Evaluate the average of $x$-coordinate:
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \frac{3}{4}\mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} x{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{4}\mathop \smallint \limits_{x = - 1}^1 x\left( {y|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = \frac{3}{4}\mathop \smallint \limits_{x = - 1}^1 \left( {x - {x^3}} \right){\rm{d}}x$
$ = \frac{3}{4} \left( {\frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right)|_{ - 1}^1 =\frac{3}{4} \left( {\frac{1}{2} - \frac{1}{4} - \frac{1}{2} + \frac{1}{4}} \right) = 0$
Evaluate the average of $y$-coordinate:
$\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \frac{3}{4}\mathop \smallint \limits_{x = - 1}^1 \mathop \smallint \limits_{y = 0}^{1 - {x^2}} y{\rm{d}}y{\rm{d}}x$
$ = \frac{3}{8}\mathop \smallint \limits_{x = - 1}^1 \left( {{y^2}|_0^{1 - {x^2}}} \right){\rm{d}}x$
$ = \frac{3}{8}\mathop \smallint \limits_{x = - 1}^1 \left( {1 - 2{x^2} + {x^4}} \right){\rm{d}}x$
$ = \frac{3}{8}\left( {x - \frac{2}{3}{x^3} + \frac{1}{5}{x^5}} \right)|_{ - 1}^1 = \frac{3}{8}\left( {1 - \frac{2}{3} + \frac{1}{5} + 1 - \frac{2}{3} + \frac{1}{5}} \right) = \frac{2}{5}$
So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {0,\frac{2}{5}} \right)$.
