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Apply divergence Gauss to find $\iint_S FdS$ where $F=(4x^3,e^{x^4}-y,z)$ and $ S$ is the portion of the paraboloid $z=x^2+y^2-6$ below $z=3$ oriented so that unit vector $\mathbf{n}\cdot(0,0,1)<0$. Hint: Remember $\int_0^{2\pi}\cos^2 xdx=\pi$.

I'm asked to solve this problem. I already computed div adding up the partial derivatives $$div(F)=12x^2 $$ So Now I have the Identity $$\iint_S F\cdot ndS = \iiint_B div(F)dV $$ But I'm not sure how to parametrize the region given to transform into $B$ (any help to continue?):

enter image description here

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  • $\begingroup$ To parametrize $S$ use $(r,\varphi)\mapsto (r\cos(\varphi), r\sin(\varphi), r^2-6)$ for $r\in [0,3]$. If you don't use Stoke's theorem, you don't have to calculate curl of $F$. $\endgroup$
    – Jochen
    Jul 24 at 13:37
  • $\begingroup$ Why did you calculate a curl? There was no curl in the surface integral in the problem, nor were you asked to compute a line integral that would necessitate using Stokes' theorem. This looks like a straightforward set up the surface integral and integrate, unless you would like to use Divergence theorem. $\endgroup$ Jul 24 at 13:39
  • $\begingroup$ @NinadMunshi I'm asked to use divergence Gauss I thought It was necessary. $\endgroup$
    – Valent
    Jul 24 at 13:40
  • $\begingroup$ This seems like a weird problem for the divergence theorem, because you'll need to pick a solid and then subtract off the integrals along the other parts of the boundary to go that way. Are you sure you read the statement correctly? $\endgroup$
    – Ian
    Jul 24 at 13:45
  • $\begingroup$ @Ian Yes it says apply Gauss to find the integral $\iint_S FdS$. Could please bring some hints? $\endgroup$
    – Valent
    Jul 24 at 13:48
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I think the graph is not right. And the divergence should be $$ \text{div}(F)=12x^2\,. $$

The key problem of this question is that you can't find $B$ such that $S=\partial B$. However, you can add another boundary to make this happen.

Set $x=r\cos(\theta),y=r\sin(\theta),z=r^2-6$. Now, define the union of paraboloid and circle $C=\{(x,y,z)|x^2+y^2=9,\ z=3\}$: $$ S'= S\cup \{(x,y,z)|x^2+y^2=9,\ z=3\}\,, $$ where vector $n$ points outwards. Then $$ \int\int_{S}F\cdot n\text{d}S=\int\int_{S'}F\cdot n\text{d}S'-\int\int_{C}F\cdot n\text{d}C=\int\int_{S'}F\cdot n\text{d}S'-27\pi. $$ Denote $$ B=\{(x,y,z)|x^2+y^2\leq z+6,\ -6\leq z\leq 3\}\,, $$ then $$ S'=\partial B\,. $$ Then we use divergence theorem to calculate $$ \begin{aligned} \int\int_{S'}F\cdot n\text{d}S'&=\int\int\int_{B}\text{div}(F)\text{d}V =\int^3_{-6}\int^{2\pi}_0\int^{\sqrt{z+6}}_012r^3\cos^2(\theta)\text{d}r\text{d}\theta\text{d}z\\ &=\int^3_{-6}\int^{2\pi}_0\int^{\sqrt{z+6}}_012r^3\cos^2(\theta)\text{d}r\text{d}\theta\text{d}z=\int^{3}_{-6}3(z+6)^{2}\text{d}z\int ^{2\pi}_0\cos^2(\theta)\text{d}\theta\\ &=9^3\times\pi=729\pi \end{aligned} $$ which implies $$ \int\int_{S'}F\cdot n\text{d}S'=729\pi-27\pi=702\pi\,. $$

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