Stokes' Theorem for Line Integrals on Closed Curves: A Problem Solution

[bigevil]

Homework Statement

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 $$x^2 + y^2 + z^2 = 2(x+y)$$

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 $$-2\sqrt{2}\pi$$

 

[HallsofIvy]

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.

 

[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.

 

[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.

 

[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.