evaluate this integral using gauss's divergence theorem

Question

$$\iint\bar N \cdot\bar F ds$$ where $\bar F=4xi+3yj-2zk$ and $S$ is the surface bounded by $x=0, y=0,z=0$ and $2x+2y+z=4$. I solved this question $\nabla\cdot\bar F=4+3-2=5$but the $$\iiint\nabla\cdot\bar Fdv=\iiint5dxdydz$$ but what will be limits of that triple integration?

Answer 1

The volume is bounded by the given four planes $$ x = 0 \\ y = 0 \\ z = 0 \\ 2x + 2y + z = 4 $$ thus a tetraeder with vertices $(0,0,0), (2,0,0), (0,2,0), (0,0,4)$.

You have to decide some parametrization for it.

E.g. $z$ varying from $0$ to $4$, $y$ from $0$ to $(4-z)/2$, $x$ from $0$ to $(4-2y-z)/2$.

Answer 2

There are two facts all vector calculus students should know. 1. The volume of a (generalized) cylinder is area of the base times the perpendicular height. 2. The volume of a (generalized) cone is $1/3$ area of the base times the perpendicular height.

The volume in your question is a generalized cone. The base is a triangle of area $2$. The height is $4$. So the volume is $8/3.$

The value of your triple integral is, then, $5 \times 8/3 = 40/3.$