# Use The Divergence Theorem to evaluate the flux

Let $\vec{F}(x, y, z) = (\sin z + xy^2)\vec{i} + x^2e^{3z}\vec{j} + (\cos ^3x + x^2z)\vec{k}.$

Let $T$ be the surface bounding the region $R$ given by $x^2 + y^2 \leq z \leq 6 - \sqrt{x^2 + y^2},$ oriented outward. Use the Divergence Theorem to evaluate the flux: $$\iint_{T} \vec{F}.d\vec{S}$$

I am having trouble applying the Divergence Theorem to calculate the flux, I got $$\operatorname{div}\vec{F} = \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) \vec F = y^2 + 0 + x^2$$ but then I don't know how to proceed.

## 1 Answer

Switch to polar coordinates to do the triple integral:

$x=r\cos \theta ;\ x=r\sin \theta ;\ z=z$.

Then, applying the Divergence Theorem amounts to evaluating

$\int_{0}^{2\pi }\int_{0}^{6}\int_{r^{2}}^{6-r}r^{3}dzdrd\theta .$

• The integrand is $r^3$. – Kuifje Dec 16 '15 at 23:31
• yes it is thanks – Matematleta Dec 16 '15 at 23:43
• Why is $r$ in $dr$ between 0 and 6 ? – NPLS Aug 5 '18 at 19:33