# Triple integral question

In a textbook problem, I am asked to find the flux through a given surface using the divergence theorem. That is, $$\iiint_{G} \nabla \cdot \vec{F} \, dV = \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS$$

Given $\vec{F} = x^3\hat{i} + y^3\hat{j} + z^3\hat{k}$, find the flux through the cylinder $x^2 + y^2 = 4$ lying below $z=3$ and above $z=0$.

My solution

Simply, $\nabla \cdot \vec{F} = 3x^2\hat{i} + 3y^2\hat{j} + 3z^2\hat{k}$. Substituting this into the integral, we obtain

\begin{align*} \iiint_{G} \nabla \cdot \vec{F} \, dV &= \iiint_{G} 3x^2\hat{i} + 3y^2\hat{j} + 3z^2\hat{k} \, dV \\ &= \int_{0}^{2 \pi} \int_{0}^{2} \int_{0}^{3} 3z^2 + 3r^2 dz dr d\theta \\ &\vdots \\ &= 156 \pi \end{align*}

however, the book gives a solution of $180 \pi$. Why is this?

• I've edited the array; please make sure it still says what you intended. – user61527 Mar 30 '14 at 23:27
• You are missing a $r$ from the Jacobian. – Mark Fantini Mar 30 '14 at 23:36

You don't have the correct volume form for cylindrical coordinates: You should integrate against $r dr d\theta dz$, similar to polar coordinates. (That is, the volume form $dx dy dz$ is given by $r dr d\theta dz$ in cylindrical coordinates, because the Jacobian is $r$.) Then the integral becomes
$$\int_0^{2\pi} \int_0^2 \int_0^3 3z^2 r + 3r^3 dz dr d\theta$$
This indeed gives $180 \pi$.