Divergence theorem(calculating flux of vector field)

Question

I need to find flux of vector field $F=(x,y,z)$ over area $T=\{|x|+|y|+|z|\le 4|(x,y,z)\in \mathbb R^3\}$ oriented in the direction of outside normal, using:a)divergence theorem; b)directly. First, using divergence theorem i get integral $\iiint 3dV$. Then, for the bounds, $z$ should always be in $[0,4]$. In $OXY$ plane I get square which contains two symmetric triangles. Then I thought to find bounds for the triangle in first and fourth quadrant like $0\le x\le 4, x-4\le y\le 4-x$(and then multiply integral with $2$). And what is left is to integrate. Is this correct? For the second part I should calculate integral $F\cdot n_0$ where $n_0$ is normal vector, but to find it, first I need to calculate gradient which I guess won't be that easy because of absolute values.

Answer 1

Yes you have correctly applied the divergence theorem. Now it is about finding the volume of the region.

See the base of the pyramid in the below diagram, which is in $XY$ plane. Now there are two square pyramids, one above $XY$ plane and one below with vertex at $(0,0,4)$ and $(0,0,-4)$.

Side of the base $ b = 4\sqrt2$ and height is $h = 4$. You can simply use the formula for pyramid volume $V = \frac{1}{3}b^2h$. Multiply by $2$ as you have two pyramids.

If you are going integral route, take the base above $x-$axis and that part of the pyramid for positive $z$. Find volume of it and then multiply by $4$.

$V = 4 \displaystyle \int_0^4 \int_{0}^{4-z} \int_{y+z-4}^{4-y-z} \ dx \ dy \ dz$