Solution
= Solution
{c}
The region is the part of the disk $x^2+y^2\leq a^2$ in the first quadrant above $y=x$. In <polar coordinates>, it is $0\leq r\leq a$ and $\pi/4\leq\theta\leq\pi/2$. The integrand becomes $r$, and the area <Jacobian determinant> contributes another $r$. Thus
$$
\boxed{\int_{\pi/4}^{\pi/2}\int_0^a r^2\,dr\,d\theta=\frac{\pi a^3}{12}.}
$$