Let $G=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1 , \quad 0\leq z\leq 1\}$
Let $f: \mathbb R^3\to\mathbb R^3,\quad f(x,y,z)=\begin{pmatrix}yz^2\\-x\\ye^z\end{pmatrix}$
Calculate $\int_M curl(f)\cdot n dS$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes: The boundary consists of the bottom and the cap's one. We parametrize both:
$\gamma_B [2\pi,0]\to\mathbb R^3, \quad t\mapsto \begin{pmatrix}\cos(t)\\ \sin(t) \\ 0\end{pmatrix} \qquad \dot{\gamma_B}=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}$
$\gamma_C [0,2\pi]\to\mathbb R^3, \quad t\mapsto \begin{pmatrix}\cos(t)\\ \sin(t) \\ 1\end{pmatrix} \qquad \dot{\gamma_C}=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}$
Now using stokes theorem we get
$\int_M curl(f)\cdot n dS=\int_{\gamma_B+\gamma_C=\partial M}f ds$
$=\int_{2\pi}^0\begin{pmatrix}0\\-\cos(t)\\ \sin(t)\end{pmatrix}\cdot\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}dt+\int_0^{2\pi}\begin{pmatrix}\sin(t)\\-\cos(t)\\ \sin(t)\end{pmatrix}\cdot\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}dt$
$=\int_{2\pi}^0-\cos^2(t)dt+\int_0^{2\pi}-\sin^2(t)dt-\cos^2(t)dt=\pi-2\pi=-\pi$
Directly:
We parametrize the surface of $M$.
$\Phi:[0,2\pi]\times [0,1]\to\mathbb R^3, \quad (t,z)\mapsto \begin{pmatrix}\cos(t)\\ \sin(t)\\ z\end{pmatrix}$
$\Phi_t\times \Phi_z=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}\times \begin{pmatrix}0\\0\\ 1\end{pmatrix}=\begin{pmatrix}\cos(t) \\ \sin(t) \\ 0\end{pmatrix}$
We see that $\Phi_t\times \Phi_z$ is pointing in the correct direction. We calcualte the curl:
$curl(f)=\begin{pmatrix}\partial_x \\ \partial_y \\ \partial_z\end{pmatrix}\times\begin{pmatrix}yz^2\\-x\\ye^z\end{pmatrix} = \begin{pmatrix}e^z \\ 2yz \\ -1-z^2\end{pmatrix}$
$\int_M curl(f)\cdot n dS=\int_0^{2\pi}\int_0^1 \begin{pmatrix}e^z \\ 2\sin(t)z \\ -1-z^2\end{pmatrix} \cdot \begin{pmatrix}\cos(t) \\ \sin(t) \\ 0\end{pmatrix} dzdt$
$=\int_0^{2\pi} \int_0^1 e^z\cos(t)+2\sin^2(t)zdzdt$
$=\underbrace{\int_0^{2\pi}\cos(t)dt}_{=0}\int_0^1 e^z dz + 2\int_0^{2\pi}\sin^2(t)dt\int_0^1zdz=0+\frac{1}{2}2\pi=\pi$
Question: So as you can see, the two results don't match up and I'm not sure why.
I think that in order to campute the flow from inside to outside you should take the bottom circle $\gamma_B$ counterclockwise and the cap circle $\gamma_C$ clockwise: $$=\int^{2\pi}_0-\cos^2(t)dt+\int^0_{2\pi}-\sin^2(t)dt-\cos^2(t)dt=-\pi+2\pi=\pi.$$ Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points along the normal of the surface, and the left hand is over the surface.