How to evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$? Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$
Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side.
How we can solve such integrals?I need your help.
Without using the Gauss Divergence theorem.
 A: First parametrize the surfaces $S_1$ (top), $S_2$ (bottom), and $S_3$ (side) as $\bf{x}=\bf{x}(s,t)$ and then use the following formula:
$$ \iint_S \bf{F}\cdot dS=\iint_T\bf{F}(\bf{x}(s,t))\cdot \left(\frac{\partial \bf{x}(s,t)}{\partial s}\times \frac{\partial \bf{x}(s,t)}{\partial t}\right)dsdt. $$
$\bf{Update 1.}$ In fact, on $S_1$, $x=r\cos t,y=r\sin t,z=2$, $t\in[0,2\pi],r\in[0,2]$, then
$$\frac{\partial \bf{x}(r,t)}{\partial r}\times \frac{\partial \bf{x}(r,t)}{\partial r}=\{0,0,r\}, \bf{F}(\bf{x}(s,t))\cdot \left(\frac{\partial \bf{x}(s,t)}{\partial s}\times \frac{\partial \bf{x}(s,t)}{\partial t}\right)=r^3$$
and hence
$$ I_1=\iint_{S_1} \bf{F}\cdot dS=\int_0^{2\pi}\int_0^2r^3drdt=8\pi. $$
Similarly, on $S_2$, $x=r\cos t,y=r\sin t,z=0$, $t\in[0,2\pi],r\in[0,4]$, then
$$-\frac{\partial \bf{x}(r,t)}{\partial r}\times \frac{\partial \bf{x}(r,t)}{\partial r}=-\{0,0,r\}, \bf{F}(\bf{x}(s,t))\cdot \left(\frac{\partial \bf{x}(s,t)}{\partial s}\times \frac{\partial \bf{x}(s,t)}{\partial t}\right)=-r^3$$
and hence
$$ I_2=-\iint_{S_3} \bf{F}\cdot dS=-\int_0^{2\pi}\int_0^4r^3drdt=-128\pi. $$
However, on $S_2$, $x=r\cos t, y=r\sin t, z=4-r$, $t\in[0,2\pi],r\in[2,4]$, then
$$\frac{\partial \bf{x}(r,t)}{\partial r}\times \frac{\partial \bf{x}(r,t)}{\partial r}=\{r\cos t,r\sin t,r\}$$
and 
$$\bf{F}(\bf{x}(s,t))\cdot \left(\frac{\partial \bf{x}(s,t)}{\partial s}\times \frac{\partial \bf{x}(s,t)}{\partial t}\right)=4r^2.$$
Thus
$$ I_3=\iint_{S_2} \bf{F}\cdot dS=\int_0^{2\pi}\int_2^4r^2drdt=\frac{448\pi}{3}. $$
So
$$ \iint_{S} \bf{F}\cdot dS=I_1+I_2+I_3=\frac{88\pi}{3}. $$
$\bf{Update 2.}$ If you use the divergence theorem, then
$$\iint_S \bf{F}\cdot dS=\iiint_{V}\text{div}FdV=2\iiint_VzdV$$
where $V$ is the volume enclosed by $S_1,S_2,S_3$,and easy calculation shows that the answer is $\frac{88\pi}{3}$.
