Answer
$\dfrac{16r^3}{3}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $$\ Volume =\iint_{D} f(x,y) \space dA$$
We have: $D$ is a circle having radius $r$ with center at the origin and the region $D$ can be defined as:
$D=\left\{ (x, y) | -r \leq x \leq r, \ -\sqrt {r^2-x^2} \leq y \leq \sqrt {r^2-x^2} \right\}$
Now, $V=\iint_{D} f(x,y) \ dA \\=2 \int_{-r}^{r} \int_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } \sqrt {r^2-y^2} dy \ dx \\ = 2\int_{-r}^{r} [\dfrac{r}{2} \sqrt {r^2-y^2}+\dfrac{r^2}{2} \sin^{-1} (\dfrac{y}{r}) ]_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } dx \\ =4 \times \int_{0}^{r}x \sqrt {r^2-x^2} +r^2 \cos^{-1} (x/r) \ dx \\=\dfrac{4}{3} r^3+4r^3\\=\dfrac{16r^3}{3}$
