Given the vector field:
$F = \langle z, y, zx \rangle$ and the tetrahedron enclosed by the coordinate planes and the plane
$\frac{x}{4}+\frac{y}{2}+\frac{z}{5}=1 \iff z = 5-\frac{5x}{4}-\frac{5y}{2}$
We want the flux through the surface of this tetrahedron.
We thus know that, by the divergence theorem:
$\int\int_SF\cdot ds = \int\int\int_Vdiv(F)dV$
We can choose our bounds as:
$0\leq x\leq4;$
$0\leq y \leq 5-\frac{5x}{4};$
$0\leq z \leq 5 – \frac{5x}{4}-\frac{5y}{2}$
And we know:
$div(F) = 0+1+x =1+x$
Thus we get the integral:
$\int^{4}_0\int^{5-\frac{5x}{4}}_0\int^{5-\frac{5x}{4}-\frac{5y}{2}}_0 (1+x)dzdydx = -50/3$
This is in fact incorrect, but I do not know why nor what the right answer is
Best Answer
Your bounds for $y$ are incorrect. The intersection of the plane and the $xy$-plane occurs when $z = 0$: $$\frac{x}{4} + \frac{y}{2} = 1 \iff y = 2 - \frac{x}{2}$$
So your bounds for $y$ should be $0 \leq y \leq 2 - \frac{x}{2}$.