# Verify Stokes' Theorem for this vector field on a surface

Elder1994
Homework Statement:
Hello, my H.W states: Verify the Stokes Theorem for the vector field A=[(4x/3 - 2y), (3y - x), 0] and the surface F:= {(x/3)^2 + (y/2)^2 =< 1, z = 0}
Relevant Equations:
∫∫∇×A⋅dS=∫A⋅dr
I do not understand how can I parameterize the surface and area and line differentials.

## Answers and Replies

Homework Helper
You must post some try or thought for us to help you. In any case, I would say that parametrize a surface or a line is to assign one or more parameters to each point on the surface or the line, for example, imagine the parabola defined in the ##x-y## plane by the equation
$$y=\frac{x(1-x)}{2}$$ you can parametrize it like $$y=-1+3t-2t^2, x=2-2t, \qquad \text{ with } -\infty < t <\infty$$
Notice that ##t## is a parameter, and that, for any choice of ##t## you get a point ##(x,y)## that lies in the desired parabola. In general, to parametrize a curve you will need one parameter and to parametrize a surface you will need two parameters.

Post some tries and we will be able to help you more.

berkeman
Homework Helper
Gold Member
What did you get for ##\nabla \times \vec A##? Think about parameterizing the surface with ##x = 3\cos\theta,~y=2\sin\theta## and similarly for the boundary curve.

Eclair_de_XII
Think about parameterizing the surface with ##x = 3\cos\theta,~y=2\sin\theta## and similarly for the boundary curve.

Would this not be the parameterization for the boundary and not the surface?

vanhees71
Elder1994
Hello, I've read all your comments, thanks for answering! I've tried to solve the problem and for the line integral, I used de parametrization: ## x = 3\cos\theta, ~y=2\sin\theta ##. I also defined my ##d\vec r## as ##(-\sin\theta , \cos\theta , 0) d\theta##, after this the line integral (from 0 to ##2\pi##) gave me ## \pi## as result. Then for the surface integral the ## \nabla \times \vec A ## , I obtained ## (0,0,1)## and the normal vector ##(\frac {\partial r(u,v)} {\partial u} \times \frac {\partial r(u,v} {\partial v})## ( where ##r(u,v) = (u,v,0)##) is ##(0,0,1)## as well. With all this, and using a change of variables my surface integral is: ## \int_{0}^{2\pi} \int_{0}^1 r \, dr \, d\theta##
This way, both results are ##\pi##, but I'm not sure if I chose the right limits and the right parametization.

Eclair_de_XII
I also defined my ##d\vec r## as ##(-\sin\theta , \cos\theta , 0) d\theta##, after this the line integral (from 0 to ##2\pi##) gave me ##\pi## as result.

If you want the differential of the parametrization ##\vec r(\theta)## you have defined, you will have to use it:

##\vec r(\theta)=(x(\theta),y(\theta),z(\theta))##

You know what ##x(\theta)##, ##y(\theta)##, and ##z(\theta)## are because they have been given to you explicitly. If you wish to find the differential ##d\vec r##, you will have to differentiate each coordinate of the vector function with respect to your independent variable ##\theta##:

##\frac{d\vec r}{d\theta}=(\frac{d}{d\theta}x(\theta),\frac{d}{d\theta}y(\theta),\frac{d}{d\theta}z(\theta))##

This is your ##d\vec r##; just "append" ##d\theta## to the right sides of the entire expression and "cancel" them out when necessary. Someone can explain this better than me, I am sure.

##r(u,v) = (u,v,0)##

This is not the parametrization of the ellipse described in the problem:

##F=\{(x,y,z)\in \mathbb{R}^3: (\frac{x}{3})^2+(\frac{y}{2})^2\leq 1, z=0\}##

Try to formulate one using polar coordinates. It should be very similar to the parametrization for the boundary given earlier, as already stated. Try drawing a picture of the thing, because it will help. You should not need to draw the whole 3-D space, by the way.

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