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I really would appreciate it if someone could help me with this problem below that I am having trouble with. The problem that I am trying to ask is down below, hyperlinked. Thank you.

Here is what I have done so far with the problem.

$S_1$ can be parametrized into $r(\theta,z) = <sin\theta, cos\theta, z>$. The Jacobian transform gives $r_{\theta} \times r_{z} = \, <cos\theta ,sin\theta ,0>$

$S_2$ can be parametrized into $r(\phi,\theta)=<sin\phi\, cos\theta, sin\phi\,sin\theta, cos\phi+1>$. The Jacobian transform gives $r_{\phi} \times r_{\theta} = \, <sin^2\phi cos\theta ,-(sin\phi)^2sin\theta ,2sin\phi cos\theta>$.

$$S_1:\int_0^{2\pi}\int_0^1F(r(\theta,z))\cdot<cos\theta,sin\theta,0>\,dzd\theta$$ $$S_2:\int_0^{2\pi}\int_0^{\pi/2}F(r(\phi,\theta))\cdot<sin^2\phi cos\theta ,-(sin\phi)^2sin\theta ,2sin\phi cos\theta>\,d\phi d\theta$$ After calculating the two double integrals for $S_1$ and $S_2$, adding the two values should give the answer, is that right? I really would appreciate if someone could help answer my question, even if just briefly. Thank you.

Click here to see Problem

zzzl
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  • The normal vector in the integral for $S_2$ looks dubious. There is a simpler way of getting it: $S_2$ is the upper hemisphere around $(0,0,1)$ who's normal vectors in Cartesian coordinates are $2(x,y,z-1),.$ Now use polar coordinates. This might be simpler to practice those things. Your $F$ looks more complicated than necessary for teaching the essentials on surface integrals. – Kurt G. Jun 21 '23 at 13:43
  • Thank you very much sir. Yes I have also felt the calculations were a bit unnecessarily complicated. If I take the normal vector $2(x,y,z-1)$ and use polar coordinates, am I ending up with a double integral of $2F(\theta,z)\cdot(cos\theta,sin\theta,z-1)$? – zzzl Jun 21 '23 at 18:45
  • I was wondering about this factor $2$ myself. Maybe it should not be there. As explained in the link the correct normal is your cross product $r_\phi\times r_\theta,.$ However you can use the polar coordinate version of $(x,y,z-1)$ to sanity check if you calculated that cross product correctly. – Kurt G. Jun 21 '23 at 18:48
  • Yep. Thank you sir. – zzzl Jun 21 '23 at 18:53
  • You can also close the surface by glueing $x^2+y^2\leq 1,z=0$ and then using divergence theorem. Have you considered this approach? – Matthew H. Jul 09 '23 at 04:39

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