# Flux of $\vec F=\langle x^2,-y^2,z^2\rangle$ over region between two parallel planes in the first octant

Use the divergence theorem to compute the net outward flux of $$\vec F(x,y,z)=x^2\,\vec\imath-y^2\,\vec\jmath+z^2\,\vec k$$ over the region $$D$$, where $$D$$ is the space between the planes $$z=3-x-y$$ and $$z=9-x-y$$ in the first octant.

Included is a plot of the region $$D$$.

By the divergence theorem, I know the flux is given by

$$2\iiint_D(x-y+z)\,\mathrm dx\,\mathrm dy\,\mathrm dz$$

and I can compute the integral easily enough (its value is $$540$$ by my calculation, confirmed both via divergence theorem and integration over each face), but I am wondering if there is a change of coordinates that I can employ to make the computation even easier. My instinct would be to transform $$D$$ into another triangular prism with uniform dimensions such that e.g. the triangular faces of the prism have the same area. Possibly something like

$$\begin{cases}x(u,v,w)=u\\y(u,v,w)=v\\z(u,v,w)=w-u-v\end{cases}$$

so that $$3\le w\le9$$ and the Jacobian $$J$$ is such that $$|\det J|=1$$, but I'm not sure what the other variables' bounds would be. Perhaps this change of variables is incorrect. Is there some way to make this work?

• The answer is 1080 ? I have solved the integral using cartesian form. Limits as $3\leq x \leq 9$ $3-x \leq 9-x$ $3-x-y \leq 9-x-y$ when integrating it gets simplify over and over. – Vedant Chourey May 18 at 4:51
• @VedantChourey The outward flux over the triangular faces is $-\frac{27}4$ in the plane $z=3-x-y$ and $\frac{2187}4$ in the plane $z=9=x=y$, giving a net flux of $\frac{2160}4=540$. The remaining three faces contribute nothing. – user170231 May 19 at 20:14
• In that case we have to calculate the net flux through th reiangular region. – Vedant Chourey May 20 at 3:09