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 defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$
We have: $D$ is a circle having a radius $r$ with center at the origin.
Thus, we can express the region $D$ as follows:
$$D=\left\{ (x, y) | -r \leq x \leq r, \ -\sqrt {r^2-x^2} \leq y \leq \sqrt {r^2-x^2} \right\}
$$
Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\=2 \int_{-r}^{r} \int_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } \sqrt {r^2-y^2} dy \ dx \\ = 2\int_{-r}^{r} [\dfrac{r}{2} \sqrt {r^2-y^2}+\dfrac{r^2}{2} \arcsin (y/r) ]_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } dx \\ =4 \int_{0}^{r}x \sqrt {r^2-x^2} +r^2 \arccos (x/r) \ dx \\=\dfrac{4r^3}{3}+4r^3\\=\dfrac{16r^3}{3}$$