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Consider the surfaces $S_1=\{(x,y,z\in\mathbb{R^3}):x^2+y^2=9-z, \ z\geq0 \}$ and $S_2=\{(x,y,z)\in\mathbb{R^3}:x^2+y^2\leq9,\ z=0\}$ and the vector field $F=(y,2z,-3y^2)$. I know that by stokes theorem I have that

$$ \int\int_{s_1}\nabla \times F\cdot dS_1=\int\int_{s_2}\nabla\times F\cdot dS_2 $$ My problem is the following: if I take the parameterizations $\Phi_1=(\cos(\theta),\sin(\theta),z)\ \ 0\leq\theta\leq2\pi\ \ 0\leq z\leq9$ for the surface $S_1$ and $\Phi_2=(r\cos(\theta),r\sin(\theta),0)\ \ 0\leq\theta\leq2\pi\ \ 0\leq r\leq3$ for $S_2$ I obtain that $$ \int\int_{s_1}\nabla \times F\cdot dS_1=0 $$ $$ \int\int_{s_2}\nabla\times F\cdot dS_2=-9\pi $$ And this must be impossible. I don't know where is my mistake, can someone explain me?.

In my calculations a get that $\nabla\times F=(-6y-2,0,-1),\ J(\Phi_1)=(\cos(\theta),\sin(\theta),0), \ J(\Phi_2)=(0,0,r)$.

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  • $\begingroup$ I think your issue is with the parametrization of $\Phi_1$ $\endgroup$ – Andrei Oct 5 '18 at 17:20
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Your $\Phi_1$ parametrization is going to produce a cylinder of radius 1 and height 9.

Instead, write the equation for $S_1$ as $z = 9-x^2-y^2$. Then you can parametrize it as $$ \Phi_1(r,\theta) = (r \cos\theta, r \sin\theta,9-r^2) $$ for $0 \leq \theta \leq 2\pi$, $0 \leq r \leq 3$.

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  • $\begingroup$ Yes, you have reason! $\endgroup$ – Gödel Oct 5 '18 at 17:22

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