# Divergence Theorem to compute volume bounded by paraboloid

Evaluate the integral $$\int_V \nabla \cdot \underline{r}\,dV$$ where $V$ is bounded by the surface $S_c$($S_c$ = part of the surface $z = a^2 - x^2 - y^2$ for which $z \geq 0$) and the plane $z=0$.

Attempt: I have done it via surface integral and obtained the correct answer but when I do it by computing the above, I get something different.

$\nabla \cdot r = 3$ (special case when we take the divergence of a position vector) so we just compute $$3 \int_V dV = 3 \int_0^a \int_0^{2 \pi} \int_0^{\pi/2} r^2 \sin \theta d\theta d\phi dr.$$ Evaluating does not give the required $3 \pi a^4/2$ obtained in the surface integral calculation.

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You need to use cylindrical coordinates, as the surface has cylindrical symmetry. The integral you want is

$$3 \int_0^{a^2} dz \: \int_0^{a^2-z} dr \,r \: \int_0^{2 \pi} d\theta$$

Note that the volume element in cylindrical coordinates is $r\,dr\,d\theta\,dz$, and the bounds are explicitly set by the equation of the paraboloid. The $3$ comes from your divergence. You should be able to do out this integral.

My answer for this integral is $3 \pi a^4/2$.

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Do I have to do the integral in cylindrical coords? The rest of the question (i.e the part where I had to do the surface integral I did via spherical coords as outlined in the question). That said, I see my integral that I set up does not distinguish between a hemisphere and a paraboloid. –  CAF Mar 20 '13 at 9:39
Should they not be the same? The $3\pi a^4/2$ was a show that... so that is how I know I did that part correct. –  CAF Mar 20 '13 at 9:43
Why do you have to do it in spherical coordinates for a surface with cylindrical symmetry? To assign such a problem that way is not a good way to teach spherical coordinates as nobody in their right mind would compute a volume over such a region that way. A paraboloid as you define here has an axis of symmetry. –  Ron Gordon Mar 20 '13 at 9:43
This is the volume I get for this region. I do not know how you got your other result. I would make sure you posed the problem properly. –  Ron Gordon Mar 20 '13 at 9:45
The parametrisation we were given in the question was $r = asin \theta cos \phi e_1 + a \sin \theta \sin \phi e_2 + a^2 \cos^2 \theta e_3$ –  CAF Mar 20 '13 at 9:45