# Stokes theorem problem

1. Feb 23, 2009

### bigevil

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

Please help me to check whether I did the right working for this problem. Thanks. The numerical answer is correct but I'm not very sure if the working is correct also.

Find $$\int y dx + z dy + x dz$$ over the closed curve C which is the intersection of the surfaces whose equations are $$x + y = 2$$ and $$x^2 + y^2 + z^2 = 2(x+y)$$

3. The attempt at a solution

First, I note that the integral required is the line integral for F . dr where F = (y, z, x). Since the curve is closed, we can apply Stokes' theorem. By Stokes' theorem $$\int F . dr = \iint (curl F) \cdot \b{n} dA$$.

Curl F = (-1,-1,-1) after applying the cross product.

Then I sketch the surfaces on the x-y axis and pick out the normal vector $$n = \frac{1}{\sqrt{2}}(\b{i} + \b{j})$$. Also $$\iint dA = \pi r^2$$. Then the answer for the line integral is $$-2\sqrt{2}\pi$$

2. Feb 23, 2009

### HallsofIvy

Staff Emeritus
What surface, exactly, are you integrating over? You are told that the curve is the intersection of two surfaces. To use Stoke's theorem, you must integrate over the entire surface: both of the given surfaces.

3. Feb 23, 2009

### Dick

It looks like bigevil is integrating over the plane. You don't have to integrate over both surfaces. Either one should give you the same answer. I'm a little confused at you deduced the radius of the circle. But other than that, it's fine.

4. Feb 23, 2009

### bigevil

Sorry Halls, the closed curve is the intersection of the two equations. That curve is closed and the surface it covers is a circle (of radius sqrt(2) I think) then I'm integrating over the plane.

I don't know how to describe it, but I deduced by imagining that the line slices the sphere given. (I drew the whole thing in the x-y plane, ie looking from the 'top' down.) The plane (y + x = 2) runs through the centre of the sphere. I got radius sqrt 2 for the radius of the closed surface.

Thanx Dick and Halls.

5. Feb 23, 2009

### Dick

Yes, exactly. The plane cuts through the center of a sphere with radius sqrt(2). You can use Stokes over that surface. As you did so well.