Prove a Statement about the Line Integral of a Vector Field

[Izzhov]

Homework Statement

Given the vector field \vec{v} = (-y\hat{x} + x\hat{y})/(x^2+y^2)
Show that \oint \vec{dl}\cdot\vec{v} = 2\pi\oint dl for any closed path, where dl is the line integral around the path.

Homework Equations

Stokes' Theorem: \oint_{\delta R} \vec{dl}\cdot\vec{v} = \int_R \vec{dA}\cdot(\vec{\nabla}\times\vec{v})

The Attempt at a Solution

I tried finding the curl of v, and got that the x and y components were zero, and the z component is \frac{\delta}{\delta x}(x/(x^2+y^2)) + \frac{\delta}{\delta y}(y/(x^2+y^2)), which becomes (y^2-x^2+x^2-y^2)/(x^2+y^2)^2, which is also zero. But this can't be right, because that would imply that \oint \vec{dl}\cdot\vec{v} = \int \vec{dA}\cdot0 = 0 by Stokes' Theorem, which would mean the thing I'm trying to show is false, since 2\pi\oint dl is not always zero.

 

[SammyS]

Homework Statement

Given the vector field \vec{v} = (-y\hat{x} + x\hat{y})/(x^2+y^2)
Show that \oint \vec{dl}\cdot\vec{v} = 2\pi\oint dl for any closed path, where dl is the line integral around the path.

Homework Equations

Stokes' Theorem: \oint_{\delta R} \vec{dl}\cdot\vec{v} = \int_R \vec{dA}\cdot(\vec{\nabla}\times\vec{v})

The Attempt at a Solution

I tried finding the curl of v, and got that the x and y components were zero, and the z component is \frac{\delta}{\delta x}(x/(x^2+y^2)) + \frac{\delta}{\delta y}(y/(x^2+y^2)), which becomes (y^2-x^2+x^2-y^2)/(x^2+y^2)^2, which is also zero. But this can't be right, because that would imply that \oint \vec{dl}\cdot\vec{v} = \int \vec{dA}\cdot0 = 0 by Stokes' Theorem, which would mean the thing I'm trying to show is false, since 2\pi\oint dl is not always zero.

Could it be that 2\pi\oint dl is always zero ?

 

[Izzhov]

Could it be that 2\pi\oint dl is always zero ?

I don't think it could, because \oint dl represents the length of the line being integrated over, doesn't it? And there's no reason why the length of a closed loop must always be zero.

 

[Dick]

I don't think it could, because \oint dl represents the length of the line being integrated over, doesn't it? And there's no reason why the length of a closed loop must always be zero.

Yes, it does. Your Stoke's theorem argument doesn't apply to curves that wind around the origin since the vector field is singular at the origin. But it does to curves that don't. And they don't all have arclength 0. It think the question got garbled somehow. I think your vector integral really computes a winding number around the origin. It's not related to arclength.

 

[Izzhov]

What exactly do you mean by a "winding number around the origin?"

 

[Dick]

http://en.wikipedia.org/wiki/Winding_number Your vector integral is zero if the curve doesn't contain the origin. As you checked with Stokes's theorem. It's 2pi if the curve winds once around the origin counterclockwise. It's -2pi if it winds once clockwise. Etc. Look down in the wikipedia reference to the 'Differential Geometry' section. You'll see your vector integral. I think the question they gave you is messed up.

 

[Izzhov]

I played around with the vector field and found that it does seem to hold that \oint \vec{dl}\cdot\vec{v} is equal to 2*pi times the winding number of any given closed loop, so I'm going to assume that's what my professor wanted me to prove. I can prove this is true when the winding number is zero by the divergence theorem (as you stated), but how do I go about proving it when the loop circles around the origin?

 

[Dick]

I played around with the vector field and found that it does seem to hold that \oint \vec{dl}\cdot\vec{v} is equal to 2*pi times the winding number of any given closed loop, so I'm going to assume that's what my professor wanted me to prove. I can prove this is true when the winding number is zero by the divergence theorem (as you stated), but how do I go about proving it when the loop circles around the origin?

You take a contour where the winding number is easy to compute. Say a contour along the unit circle. Now take your hard to compute contour. Draw a segment connecting the two contours. Figure out how you can run around the two contours using the segment bridging them and only enclose a region that does not contain the origin.