Find the flux of $F = x\mathbf i + z\mathbf j$ out of the tetrahedron bounded by $x + 2y + 3z = 6$ and the coordinate planes

Question

Find the flux of $F = x\mathbf i + z\mathbf j$ out of the tetrahedron $x + 2y + 3z = 6$

I realise that you would need to calculate flux for each side of the tetrahedron and then add the results together, but have no idea about how to actually go about calculating the flux, and the examples in the textbook are not especially enlightening.

Answer 1

The flux is $$\iint_SF \cdot dS$$

Hence \begin{align} \iint \langle x, z, 0 \rangle\cdot \langle -z_x,-z_y,1 \rangle dA &= \iint \bigg \langle x, \left( \frac{1}{3}\right)(6-x-2y),0 \bigg \rangle \cdot \langle 1/3,2/3,1 \rangle dA \\ &= \frac{1}{9}\iint \left( 3x + 2(6-x-2y) \right)d A \\ &=\frac{1}{9}\iint 12+x-4y\, dA \end{align}

now, you must evaluate over the region bounded by $x+2y+3z=6$ so evaluate $$\frac{1}{9}\int_0^3\,dy\int_0^{6-2y}dx\,(12+x-4y)$$ Which is left as an exercise.

Answer 2

The given plane intersects the axes at $x=6$, resp., $y=3$, resp., $z=2$. The tetrahedron $T$ therefore has volume ${\rm vol}(T)={1\over 6}\cdot 6\cdot 3\cdot 2=6$. Since ${\rm div}(F)\equiv1$ we obtain by Gauss' theorem that the flux $\Phi$ in question has value $$\Phi=1\cdot 6=6\ .$$