# Stoke's Theorem around a closed circle

1. Jul 21, 2009

### ejs12006

1. The problem statement, all variables and given/known data

What is the line Integral of the function f = yi-xj+zk (where i,j,k, are cartesian unit vectors) around a circle with radius R centered at the origin?

2. Relevant equations

Stokes Theorem: i.e. the integreal of some function between a and b is equal to the difference in the values of the antiderivitive of that function evaluated at a and b.

3. The attempt at a solution

As I understand it, the answer should be zero; since the integral is over a closed circle (and since it is a continuous function with no sigularities). In this case, a=b, so the antiderivitive of f at a minus the anti-derivitive of f at b = 0.

antiderivitive of f: F= .5(y^2)i-.5(x^2)j+.5(z^2)k

Limits of integration: a=b

Apply Stokes Theorem:F(a)-F(b) =0

However, apparently the answer is 2*pi*(R^2) (the area of a circle). No explenation is given except that stokes theorem should be used . Any Ideas? I am almost certain there is an error in the solutiuons, but I would appreciate a second opinion.

2. Jul 21, 2009

### bartek2009

Do you integrate in a clockwise direction or the usual anti-clockwise manner?
If the questions asks you to do it clockwise then I found the answer to be 2*pi*(R^2).

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Last edited by a moderator: Aug 6, 2009
3. Jul 21, 2009

### g_edgar

NO NO NO

Help stamp out this error. You refer to a theorem by somebody called Stoke. That is NOT what you want. The poor man's name was Stokes. Sir George Stokes. So...

correct: Stokes' Theorem ... or ... Stokes's Theorem

incorrect: Stoke's Theorem

Give Sir George his due!

4. Jul 21, 2009

### HallsofIvy

What you state as Stoke's theorem is NOT- it is just the fundamental theorem of Calculus.

The Stokes' theorem as given in Calculus is
$\oint_C \vec{F}\cdot d\vec{r}= \int\int_S \nabla\times \vec{F} d\vec{S}$
where C is the boundary of the surface S.

What is sometimes called Stokes' theorem in advanced itexematics, or the "generalized" Stokes' theorem is
$\int_{d\sigma} \omega= \int_\sigma d\omega$
where $\sigma$ is a simply connected manifold with boundary $d\sigma$ and $\omega$ is a differential form with differential $d\omega$.

Since this is in 3 dimensions, there are an infinite number of "circle with radius R centered at the origin". Is any information given as to which is intended?

Last edited by a moderator: Jul 24, 2009
5. Jul 24, 2009

### ejs12006

The circle is in the x-y plane, Sorry about that.

Ok it makes sense that what I've written above is just the fundamental theorem of calculus, but shouldn't it still apply? i.e. the answer of 0 is still valid?

6. Jul 24, 2009

### ejs12006

OK so I actually did it out and I get the right answer. But if anyone can explain how Stokes' theorem applies here, that would be much appreciated.

7. Jul 24, 2009

### nickmai123

Well you probably couldn't read what HallsofIvy said because his/her LaTeX wasn't in the form this forum could read.

Stokes' Theorem:

$$\int\vec{F}\cdot d\vec{r}=\int(\vec{\nabla}\times\vec{F})\cdot d\vec{S}$$

It says that instead of taking the line integral, you can change it into a flux integral of the curl(F) through the area bounded by the circle or, more generally, a curve C.

Also, the fundamental theorem does not say that F(b) - F(a) = 0. A line integral is 0 only when you integrate through a conservative vector field around a closed path. The field you provided is not conservative, so despite being a closed path, that line integral will not be 0. (To verify that, take the curl of that vector field. It is conservative if and only if the curl(F) is 0, which it is not.)

EDIT: You might want to look up the definition of a conservative vector field. I realized that I brought that up without explaining what it meant. Go to Paul's Online Math Notes. That site has a lot of answers to questions from students in maths ranging from Calcs 1 to 3, DE, and Linear Algebra.

Last edited: Jul 24, 2009
8. Jul 24, 2009

### HallsofIvy

Because this circle is in the xy-plane, z= 0 and "Stokes' Theorem" reduces to "Green's Theorem" applied to the function $y\vec{i}- x\vec{j}$.

Greens theorem says
$$\oint P(x,y)dx+ Q(x,y)dy= \int\int \left(\frac{\partial Q}{\partial x}- \frac{\partial P}{\partial y}\right) dxdy$$.

For this problem, P(x,y)= y and Q(x,y)= -x so the integral on the right is just -2 times the area of the circle.