Answer
$\displaystyle \frac{2a^{3}}{3}$
Work Step by Step
The graph crosses the x-axis for $x=0$ and for
$\sqrt{a^{2}-x^{2}}=0\Rightarrow x=\pm a$.
When x is negative, y is below the x-axis,
and for positive x, it is above the x-axis.
So, for $x\in[-a,0],\ A_{1}=-\displaystyle \int_{-a}^{0}x\sqrt{a^{2}-x^{2}}dx$
and for $x\in[0,a],\ A_{2}=\displaystyle \int_{0}^{a}x\sqrt{a^{2}-x^{2}}dx$
$A_{1}$= sub: $\displaystyle \left[\begin{array}{ll}
a^{2}-x^{2}=u, & du=-2xdx\\
x=-a\Rightarrow & u=0\\
x=0\Rightarrow & y=a^{2}
\end{array}\right]=-\int_{0}^{a^{2}}u^{1/2}(-\frac{du}{2})$
$=\displaystyle \frac{1}{2}\left[\frac{u^{3/2}}{3/2}\right]_{0}^{a^{2}}=\frac{(a^{2})^{3/2}}{3}=\frac{a^{3}}{3}$
Testing for symmetry, we find $f(-x)=-x\sqrt{a^{2}-(-x)^{2}}=-f(x)$,
so the function is odd, and the graph is symmetric about the origin.
So the area over $[0,a]$ equals $A_{1}$ (the area over $[-a,0]$ ).
Total area = $2A_{1}=\displaystyle \frac{2a^{3}}{3}$
$\displaystyle \frac{2a^{3}}{3}$
(Note that calculating $A_{2}$ with the same substitution as above leads to $A_{2}=\displaystyle \frac{a^{3}}{3}$)
