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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.

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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.

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  • $\begingroup$ If I want the flow from inside to outside, the normal vector on the bottom looks downwards and the one from the cup looks upwards, no? So the persons stnds on top of the cylinder e.g. and walks coutnerclockwise around the border of the cup. No? Basically the right-thumb rule. Like: imgur.com/a/CNGNlQo $\endgroup$
    – xotix
    Jan 26 '19 at 10:34
  • $\begingroup$ You said that the surface is a "hollow cylnder with no bottom and no cap" so why are you considering the surfaces of those caps? See page 2 here: web.mit.edu/jorloff/www/18.01a-esg/notes/topic48.pdf $\endgroup$
    – Robert Z
    Jan 26 '19 at 10:44
  • $\begingroup$ I just relaized that's exaclty my problem! It's a hollow cylinder, so my surface is the part parallel to the z-axis! So yeah, you are right. Thanks $\endgroup$
    – xotix
    Jan 26 '19 at 10:53
  • $\begingroup$ @xotix Well done!! $\endgroup$
    – Robert Z
    Jan 26 '19 at 10:59
  • $\begingroup$ 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? $\endgroup$
    – xotix
    Jan 26 '19 at 11:02

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