Evaluate the integral
$$\int _V^{ }\:\underline∇·\underline r\:dV$$
where $V$ is the volume bounded by the surface $z=a^2-x^2-y^2$ (a paraboloid) and the plane $z=0$.
How do I get started on this problem? At the moment I have $\underline r=(x,y,a^2-x^2-y^2)$ so $\underline∇·\underline r=2$, and the limits of integration are $-a\le x \le a, -\sqrt{a^2-x^2}\le y \le \sqrt{a^2-x^2}, 0\le z \le a^2-x^2-y^2$. But I don't think this is the right approach. I am asked to compare the answer to the sum $\int _{S_C}\underline r·d\underline S\:+\int _{S_B}^{ }\:\underline r·d\underline S=\frac{3\pi a^4}{2}$ where $S_C$ is the part of the paraboloid for which $z\ge 0$ and $S_B$ is the circular base of the paraboloid, which satisfies $x^2+y^2=a^2$ with $z=0$.
Best Answer
$\underline r = (x, y, z)$ and not $(x,y,a^2-x^2-y^2)$.
$z = a^2 - x^2 - y^2$ only on the paraboloid surface.
$\nabla \cdot \underline r = 3$
In polar coordinates, $0 \leq z \leq a^2 - \rho^2, 0 \leq \rho \leq a$
So the volume integral is,
$ \displaystyle \int_0^{2\pi} \int_0^{a^2} \int_0^{\sqrt{a^2 - z}} 3~\rho~ d\rho~ dz~d\theta = \frac{3 \pi a^4}{2}$