Flux of vector field $F(x,y,z) = (\frac{1}{6}x^2yz,\frac{1}{4}xy^2z,\frac{1}{12}xyz^2)$ through cube

Question

I'm trying to calculate the flux of the vector field $F(x,y,z) = (\frac{1}{6}x^2yz,\frac{1}{4}xy^2z,\frac{1}{12}xyz^2)$ through the surface: $$ \Omega = \{(x,y,z) \in \mathbb{R}^3 : 0 \leq x,y,z \leq 2\}$$ $\Omega$ is of course the cube. I tried using the divergence theorem but I did not get anywhere. I also tried to use some form of parametrization but I did not succeed.

Any hints?

Answer 1

We have

\begin{align*} \displaystyle{\int_{0}^{2}\int_{0}^{2}\int_{0}^{2}\text{div}\,F} &=\displaystyle{\int_{0}^{2}\int_{0}^{2}\int_{0}^{2}}\bigg[\frac{2}{6}+\frac{3}{6}+\frac{1}{6}\bigg]xyz\,dx\,dy\,dz \\ &= \displaystyle{\int_{0}^{2}\int_{0}^{2}\int_{0}^{2}}xyz\,dx\,dy\,dz \\ &=\displaystyle{\int_{0}^{2}\int_{0}^{2}\bigg[\frac{x^{2}}{2}}\bigg|^{2}_{0}\bigg]\,yz\,dy\,dz\\ &=\displaystyle{\int_{0}^{2}\int_{0}^{2}}2yz\,dy\,dz. \end{align*} We notice that since we have the same bounds and same variables, each individual integral will evaluate to $2$ and we will be left with $8$ as our final answer.

Answer 2

You don’t need a parametrization. Hint: your limits of integration for the triple integral you need are all from 0 to 2.