Consider this problem: $F=[4x,x^2y,-x^2z]$ and $S$ is the tetrahedron described by $(0,0,0);(1,0,0);(0,1,0);(0,0,1)$
we have to find the surface integral.
Question 1: Applying divergence theorem $\nabla\cdot F =4$ which implies the surface integral turns into $\int \int \int 4 dV$ or otherwise 4 times the volume of tetrahedron. We know volume of tetrahedron is $\sqrt{2}a^3/12$ which means the answer should be $\sqrt{2}/3$. But the answer is not that.
Question 2: If we split $dV$ into $dxdydz$ and then integrate, the limit my book says should be $0 to 1-y-z$,$0 to 1$,$0 to 1$ respectively. Can you talk a bit more about it. Intuitively I got an understanding, but how is it different the limits had it been from x+y+z=1?
Question 3: Just for the kicks I tried integrating this on the surface of the tetrahedron. I get the normal vector for only one surface, i.e the plane x+y+z =1 the others, turn into an equation where one of the variable (x,y,z) is zero [as the tetrahedron is bounded by that plane] which implies when turning into parametric form of u-v plane, the r(u,v) will be dependent on only 1 variable. For instance, one of the planes will be $x+z\leq 1$ => r(u,v)=[u,0,1-u] which implies $r_v=0$, hence cross product of $r_u$ and $r_v$ will be zero, hence the integral will be zero on that surface. This was my modus operandi. So I surface integrated over only $x+y+z \leq 1$ and I am getting a pretty way off the answer. Can someone lend me a hand here.
Thanks in advance Soham