Use the divergence theorem to calculate $\int \int F\cdot dS$ where $F=<x^3,y^3,4z^3>$ and $S$ is the sphere $x^2+y^2+z^2=25$ oriented by the outward normal.
I have found that $div(F)=<3x^2,3y^2,12z^2>$ and set up the an integral to solve using the formula $\int \int F\cdot dS=\int \int \int div(F) dv$.
As such, I converted to spherical coordinates and simplified to the expression(including the $\rho^2sin(\phi)$ factor):
$\int \int \int3\rho^4sin^3(\phi)+12\rho^4sin(\phi)cos^2(\phi)d\rho d\phi d\theta$
with the bounds$\quad0\le \rho \le 5,\quad0 \le\phi \le \pi, \quad 0\le\theta\le 2\pi$
My result came out as $47123.9$ which is apparently not correct. Anyone want to help me figure out what I'm doing wrong? Thanks!