# 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.

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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