# Calculating a flux integral using Stokes vs. directly

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.
• Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $\vec{n}$ takes care of the considered direction of the flux, right? Jan 26 '19 at 11:02