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.

 

[Gaussian97]

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.

 

[LCKurtz]

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?

 

[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.

 

[LCKurtz]

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

Yes. I inadvertently left out the $r$ going from $0$ to $1$. And for the OP's benefit, I think he should get $6\pi$ for the answer to both.

 

[Elder1994]

Hello, I reviewed the problem and I found indeed that the limits that I used were wrong, changing that the answer for both integrals is $6\pi$ ! Thank you.