Using divergence theorem, show that $\int_SF.n \,dS=7\pi/6$, where $F=xi+yj+z^2k$, and S=closed surface bounded by $x^2+y^2=z^2$ and $z=1.$
Attempt: $div\, F=2(1+z)$.
$\displaystyle \int_SF.n\, dS=2\int_V (1+z)dxdydz$
Please mention the limits and how to evaluate the integral in a simplified way.
Edit:
Tried with the following but becoming very complicated
$\displaystyle 2\,\int_{x=-1}^{1}\!\int_{y=- \sqrt{-{x}^{2}+1}}^{ \sqrt{-{x}^{2}+1}}\!
\int_{ \sqrt{{x}^{2}+{y}^{2}}}^{1}\!(1+z)\,{\rm d}z\,{\rm d}y\,{\rm d}x
$
Best Answer
Your volume of integration is the cone from $z=0$ to $z=1$, so with a cylindrical coordinates change: $$x=r\cos(\theta),\;y=r\sin(\theta),\;z=z$$ $r\in[0,z]$, $\theta\in[0, 2\pi]$ and $z\in[0, 1]$ (Do not forget the jacobian) So the integral transforms into: $$\displaystyle 2\,\int_{0}^{1}\!\int_{0}^{2\pi}\! \int_{0}^{z}\!(1+z)r\,{\rm d}r\,{\rm d}\theta\,{\rm d}z=2\pi\int_{0}^{1}(1+z)z^2 dz=2\pi(1/3+1/4)=\frac{7\pi}{6} $$