Stoke's theorem to calculate circulation

[DWill]

Homework Statement

Use the surface integral in Stoke's Theorem to calculate the circulation of the Field F around the curve C in the indicated direction.

F = (x^2)i + (2x)j + (z^2)k
C: The ellipse 4x^2 + y^2 = 4 in the xy-plane, counterclockwise when viewed from above.

Homework Equations

The Attempt at a Solution

At first I attempted to parametrize the ellipse in terms of u and v, like r(u,v) = cos(v)i + 2sin(v)j + (u)k. But I don't think this is doing it right, and also I don't know what the bounds on u would be. So I tried another way I saw done in an example:

I tried to find the normal n, which my book states to be grad(f) / |grad(f)|. Taking f to be 4x^2 + y^2 = 4, I find n = (8x i + 2y j + 0k) / sqrt(64x^2 + 4y^2). Then I try to find curl F, which is grad x F, and I find this simply = 2k. Then taking dot product of curl F and n I end up with 0, since n has no z-component and curl F has no x and y-component. This is like I expected not the right answer.

Can someone point out what is wrong here or how I do these kind of problems the correct way? Any help is appreciated.

 

[Defennder]

The question seems a little improperly stated. The line integral of the circulation of F, or curl F would be curl of the curl of F through the surface enclosed by the curve C. Or the flux of \nabla \times \nabla \times \textbf{F} through the ellipse. I would think that you might have mis-transcribed the question and meant instead the closed path integral along the ellipse.

Regardless, what you should do if you want to evaluate the surface integral is to first parametrise the surface. There's an obvious choice for a surface to be parametrised if you want to evaluate the integral easily.

There's an obvious choice for what n should be if you chose the easy surface to paramatrise. And it certainly has a z-component.

 

[HallsofIvy]

Homework Statement

Use the surface integral in Stoke's Theorem to calculate the circulation of the Field F around the curve C in the indicated direction.

F = (x^2)i + (2x)j + (z^2)k
C: The ellipse 4x^2 + y^2 = 4 in the xy-plane, counterclockwise when viewed from above.

Homework Equations

The Attempt at a Solution

At first I attempted to parametrize the ellipse in terms of u and v, like r(u,v) = cos(v)i + 2sin(v)j + (u)k. But I don't think this is doing it right, and also I don't know what the bounds on u would be. So I tried another way I saw done in an example:

I tried to find the normal n, which my book states to be grad(f) / |grad(f)|. Taking f to be 4x^2 + y^2 = 4, I find n = (8x i + 2y j + 0k) / sqrt(64x^2 + 4y^2). Then I try to find curl F, which is grad x F, and I find this simply = 2k. Then taking dot product of curl F and n I end up with 0, since n has no z-component and curl F has no x and y-component. This is like I expected not the right answer.

Can someone point out what is wrong here or how I do these kind of problems the correct way? Any help is appreciated.

Has this been edited from the original? Defennders remark about \nabla\times\nabla\times F doesn't make sense!

In any case, the problem says to use the Stokes theorem Now stokes theorem applies to any smooth curve in R³ but this problem is in the xy-plane. The normal to the surface is just \vec{k}, the unit vector in the z direction, and the surface integral is just the intgral over the ellipse. The "circulation" is just
\int_A\int \nabla\times F(x,y,z)\cdot \vec{k} dxdy
That is, you are integrating the \vec{k} component of the curl over the ellipse. You don't want to parameterize the ellipse since the whole point of Stokes theorem is to avoid having to do the path integral.

In fact, here the formula is so simple you don't really have to do any integration at all! Do you know how to find the area of an ellipse?

 

[Defennder]

Has this been edited from the original? Defennders remark about \nabla\times\nabla\times F doesn't make sense!

I interpreted the question to mean find the closed line integral of curl F along the ellipse.