The problem:

Find the flux $\textbf{F} = 3x\hat{i} + z\hat{j}$ out of the tetrahedron closed in by the plane $5x + 3y + 3z = 4$ and the xy, xz and yz planes.

My (wrong) solution:

I calculated the divergence of $\textbf{F} = 4$. Then i find the volume of the tetrahedron $$\frac{\frac{4}{5} * \frac{4}{3} * \frac{4}{3}}{3} = \frac{64}{135}$$ Then i multiply the volume by the divergence and get $\frac{256}{135}$

Any help would me much appreciated.

Update: The divergence is actually 3, and when calculating the volume I forgot to divide the base by two.

  • $\begingroup$ When you compute the volume this way, you are missing a $1/2$ factor. $\endgroup$ – Kuifje Mar 20 '17 at 17:05
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    $\begingroup$ Try calculating the flux through each of the four planes that bound the tetrahedron. $\endgroup$ – BobaFret Mar 20 '17 at 17:10

Your approach is perfectly correct. The divergence is incorrect: it is $3$. I suspect you also made an error when computing the volume the tetrahedron.

The projection of the tetrahedron in the $xy$ plane is the triangle bounded by the axes, and the line $5x+3y=4$, that is, the set $$ D \{(x,y)\;|\; 0 \le x \le \frac{4}{5}, 0 \le y \le \frac{4- 5x}{3} \} $$ There the volume equals $$ V = \iint_D \frac{4- 5x-3y}{3}\; dA = \frac{32}{135} $$ And the flux is $3V = \frac{96}{135}$.

  • $\begingroup$ The divergence was also wrong $\frac{\partial z\hat{j}}{\partial y}$ is actually 0, not 1 as I found. Thanks a lot! $\endgroup$ – Viktor Mar 20 '17 at 17:12

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