Line integral and vector fields

[Benny]

Hi, I'm having trouble with the following question.

Q. Let p be a real constant and \mathop F\limits^ \to = \left( {yz^p ,x^p z,xy^p } \right) be a vector field. For what value of p is the line integral

\int\limits_{C_2 }^{} {\mathop F\limits^ \to \bullet d\mathop s\limits^ \to } = 0

Where C_2 is any closed path in R^2.

Firstly, how can C_2 be a path in R^2 when the vector field is '3D'? That doesn't seem to make sense in the context of the line integral. Assuming that C_2 is any closed path in R^3 then it should be sufficient to find the values of p so that curl F = 0.

I found curl\mathop F\limits^ \to = \nabla \times \mathop F\limits^ \to

<br /> = \left( {x\left( {p - 1} \right)y^{p - 1} - x^p ,y\left( {p - 1} \right)z^{p - 1} - y^p ,z\left( {p - 1} \right)x^{p - 1} - z^p } \right)<br />

The answer is p = 1 but substituting p = 1 to what I found doesn't give the result curl F = 0. I've checked over my calculation a few times but I still can't see what's wrong with the curlf that I've computed. Can someone help me out?

 

[HallsofIvy]

You're right. It doesn't make sense to define a vector field in R³ and then ask you to integrate it around a curve in R²!

There are two things you could do- you assume they mean a path in the xy- plane, z= 0. Of course, that makes the problem trivial- I'm sure that is not what was intended. I would assume that R² was just a typo- that it should be R³.

You have = \left( {x\left( {p - 1} \right)y^{p - 1} - x^p ,y\left( {p - 1} \right)z^{p - 1} - y^p ,z\left( {p - 1} \right)x^{p - 1} - z^p } \right)

Good grief! You've correctly analyzed a problem in Green's theorem and messed up the "power law"?? The derivative of xⁿ is nx^n-1, not (n-1)x^n-1 as you have!

 

[Benny]

Heh, I don't know how I managed to miss my error with the power rule. Thanks for the help HallsofIvy.