Answer
$\dfrac{16r^3}{3}$
Work Step by Step
The volume under the surface is given by :$z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $\ Volume ; V=\iint_{D} f(x,y) \space dA$
Our aim is to calculate the volume of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | -r \leq x \leq r, \ -\sqrt {r^2-x^2} \leq y \leq \sqrt {r^2-x^2} \right\}$
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} (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 \cos^{-1} (x/r) \ dx \\=\dfrac{4}{3} \times r^3+4 \times r^3\\=\dfrac{16r^3}{3}$
