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.
Best Answer
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 $3V = \frac{96}{135}$.