Question

Find the flux of F = x^3 i  + y^3 j  + z^3k through the closed surface bounding the solid region x^2 + y^2 ≤ 4, 0 ≤ z ≤ 4

Answer 1

Use the divergence theorem. Let $R$ be the cylindrical region, then

$\displaystyle\iint_{\partial R}\mathbf F\cdot\mathbf n\,\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dV$

(where $\mathbf n$ denotes the unit normal vector to $\partial R$, but we don't need to worry about it now)

We have

$\mathrm{div }\mathbf F=(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial\mathbf F}{\partial x}+\dfrac{\partial\mathbf F}{\partial y}+\dfrac{\partial\mathbf F}{\partial z}$
$\nabla\cdot\mathbf F=3x^2+3y^2+3z^2$

For the solid $R$ with boundary $\partial R$, we can set up the following volume integral in cylindrical coordinates for ease:

$\displaystyle3\iiint_R(x^2+y^2+z^2)\,\mathrm dV=3\int_{z=0}^{z=4}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}(r^2+z^2)r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$
$=\displaystyle6\pi\int_{z=0}^{z=4}\int_{r=0}^{r=2}(r^3+rz^2)\,\mathrm dr\,\mathrm dz$
$=\displaystyle12\pi\int_{z=0}^{z=4}(2+z^2)\,\mathrm dz$
$=352\pi$