<s>[INST] Surface integral on one-sheeted hyperboloid I'm working on the following exercise:
"Let $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2+1,-1\leq z\leq3\}$ be a surface with orientation $\omega$.
Compute $\int\int_{\Sigma}(\nabla\times F)\cdot\omega\ dS$ knowing that $F=(-\frac{y}{3},-\frac{z}{3},-\frac{x}{3})$ and $\omega (0,1,0)=(0,1,0)$. 
Verify the result you have obtained using Stokes' Theorem."
Here's what I've done:
-> $\nabla\times F=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$;
-> I've divided the surface in three pieces:
(+) $0<z\leq 3$: parametrization $\varphi_{+}:=(x,y,\sqrt{x^2+y^2-1})\to \varphi_{+_x}=(1,0,\frac{x}{\sqrt{(x^2+y^2-1)}}), \varphi_{+_y}=(0,1,\frac{y}{\sqrt{(x^2+y^2-1)}})\to$
$\omega_+=\varphi_{+_x}\times\varphi_{+_y}=(-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)$
($0$)$\ z=0$: $\omega_0=(x,y,0)$ (ok, since $\omega (0,1,0)=(0,1,0)$;
(-)$\varphi_{-}:=(x,y,-\sqrt{x^2+y^2-1})\to \varphi_{-_x}=(1,0,-\frac{x}{\sqrt{(x^2+y^2-1)}}), \varphi_{-_y}=(0,1,-\frac{y}{\sqrt{(x^2+y^2-1)}})\to$
$\omega_-=\varphi_{-_x}\times\varphi_{-_y}=(\frac{x}{\sqrt{x^2+y^2-1}},\frac{y}{\sqrt{x^2+y^2-1}},1)$
(Notice that I've not normalized the normal vector since when computing the flux of a vector field across a surface in space we have $\int\int_S F\cdot n dS=\int\int_S F\cdot (\varphi_x \times\varphi_y)dxdy$)
So $\int\int_{\Sigma_+}(\nabla\times F)\cdot\omega_+ dS=\int\int_{\Sigma_+}(\frac{1}{3},\frac{1}{3},\frac{1}{3})\cdot (-\frac{x}{\sqrt{x^2+y^2-1}},-\frac{y}{\sqrt{x^2+y^2-1}},1)dxdy=\frac{1}{3}\int\int_{\Sigma_+}(-\frac{x}{\sqrt{x^2+y^2-1}}-\frac{y}{\sqrt{x^2+y^2-1}}+1)=?$
Here is where I get stuck, since I can't figure out what the bounds for this double integral should be; can anyone give me an hint about this?
 [/INST] By the divergence theorem, your integral equals
$$
\Phi = \iiint_E \underbrace{\nabla\cdot \nabla \times F}_{=0}\; dV-\iint_{S_1}\nabla \times F\; dS-\iint_{S_2}\nabla \times F\; dS,
$$
where $S_1$ and $S_2$ are the surfaces that close $\Sigma$ at $z=-1$ and $z=3$. $S_1$ can be parametrized as follows:
$$
\begin{cases}
x=x\\
y=y\quad \mbox{with  }\; x^2+y^2\le 2\\
z=-1
\end{cases}
$$
And similarly for $S_2$:
$$
\begin{cases}
x=x\\
y=y\quad \mbox{with  }\; x^2+y^2\le 10\\
z=3
\end{cases}
$$
It follows that
$$
\Phi = -\iint_{x^2+y^2\le 2} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\-1}\; dS -\iint_{x^2+y^2\le 10} \pmatrix{1/3\\1/3\\1/3}\cdot \pmatrix{0\\0\\1}\; dS =\frac{A(S_1)}{3}-\frac{A(S_2)}{3} = -\frac{8\pi}{3}
$$
</s>