MvCalc14

$\int_0^a[\int_{0}^{\sqrt{a^2-x^2}}xy dy]dx\,$

Now $\int_{0}^{\sqrt{a^2-x^2}}xydy\,$

= $[\frac{xy^2}{2}]_{0}^{\sqrt{a^2-x^2}}\,$

= $\frac{x(a^2-x^2)}{2}\,$

= $\frac{1}{2}[a^2x-x^2]\,$

Therefore, $I=\int_0^a[a^2x-x^2]dx\,$

= $[\frac{a^2x^2}{2}-\frac{x^3}{3}]_0^a\,$

= $\frac{a^4}{2}-\frac{a^3}{3}\,$

= $\frac{a^3}{6}[3a-2]\,$