# Stokes' Theorem Problem

## Homework Statement

Evaluate the line integral of F dot dr by evaluating the surface integral in Stokes' Theorem with an appropriate choice of S. Assume that C has a counterclockwise orientation.

F
= (y2, - z2, x); C is the circle r(t) = < 3cos(t), 4cos(t), 5sin(t) >

## Homework Equations

F[/B] is the vector field.

## The Attempt at a Solution

I found the curl of F to be <2z, -1, -2y> and I computed the line integral and got 15pi, which is the correct solution, but I have to use Stokes' Theorem to arrive at that solution. My main problem is I have no clue how to solve for the normal vector to dot with the curl of F. The main thing throwing me off is that C is already parameterized and I cannot see how to un-parameterize it to get the equation for the circle out.

Well, first off, <3cos(t),4cos(t)5sin(t)> isnt a circle-its more elliptical. You dont need to "un-parameterize" it. You need to first find a normal vector to the plane of the ellipse. (which you should use as your "capping" surface)

@DeldotB Yeah, I could not get my head around why the book was calling it a circle. Thank you though, I shall give it a go!

STEMucator
Homework Helper
So you wish to compute the line integral of the vector field using stokes theorem. Notice you have ##\vec r(t)## given as a function of ##t## only. Computing a surface integral will not be possible with the given ##\vec r(t)## as is.

If instead the surface was given in two variables, for example ##\vec r(\theta, \phi)##, then you would be able to compute a vector normal to the surface ##\vec n##. Then you would be able to use a surface integral to compute the result.

As it stands, you should evaluate the line integral using:

$$\oint_C \vec F(x, y, z) \cdot d \vec r = \int_a^b \vec F( \vec r(t) ) \cdot \vec r'(t) \space dt$$

This is Stokes' theorem, simply backwards:

$$\iint_S \text{curl}(\vec F(x, y, z)) \cdot d \vec S = \oint_C \vec F(x, y, z) \cdot d \vec r = \int_a^b \vec F( \vec r(t) ) \cdot \vec r'(t) \space dt$$

So you are technically using Stokes' theorem.

Last edited:
LCKurtz
Homework Helper
Gold Member
@BigFlorida: I do not agree with Zondrina that his argument "technically" uses Stokes' theorem. Here are some suggestions. You have ##\vec R(t) = \langle 3\cos t, 4\cos t, 5\sin t\rangle##, and you have calculated ##\nabla \times \vec F = \langle 2z, -1, -2y\rangle##. You can see right away from ##\vec R(t)## that ##4x=3y## so that tells you the curve is in that plane. Also, since your curl came out in terms of ##y## and ##z##, that suggests using them as parameters. Again, looking at ##\vec R(t)## you can see that ##\frac{y^2}{16} + \frac{z^2}{25} = 1##, so the projection of the curve on the ##yz## plane is an ellipse. You could parameterize the interior of that ellipse in the ##yz## plane in a polar coordinate like transformation $$y = 4r\cos\theta,~z = 5r\sin\theta,~0\le r \le 1,~0\le\theta\le 2\pi$$If you want to parameterize the portion of the plane inside the given curve, you just need to add the parameterization for ##x## which which you know is ##x = \frac 3 4 y = 3r\cos\theta##.

Hopefully that helps you with the parameterization of your surface. Can you take it from there?

STEMucator
Homework Helper
@BigFlorida: I do not agree with Zondrina that his argument "technically" uses Stokes' theorem.

It does 'technically' use the theorem, in another form. You can of course go the other way and use a surface, which your post has outlined.

I just wanted to say you can't use ##\vec r(t)## as is, or else you could not compute the surface integral.

LCKurtz
Homework Helper
Gold Member
It does 'technically' use the theorem, in another form. You can of course go the other way and use a surface, which your post has outlined.

I just wanted to say you can't use ##\vec r(t)## as is, or else you could not compute the surface integral.

Quoting from the original post: "Evaluate the line integral of F dot dr by evaluating the surface integral in Stokes' Theorem with an appropriate choice of S". Your post is irrelevant to that problem.

@LCKurtz Wow, thank you so much. That makes complete sense, I did not even think of projecting it onto the yz-plane. I can definitely take it from there. Thank you again!