Line integral and vector fields

In summary, the question asks for the value of p that makes the line integral of a given vector field, over any closed path in R3, equal to 0. The correct approach is to find the values of p that make the curl of the vector field equal to 0. However, in the conversation, there was confusion about whether the path should be in R2 or R3. It was determined that R3 was most likely intended and the error in using the power rule was pointed out. The final answer is p = 1, but substituting this value does not give the correct result for the curl of the vector field.
  • #1
Benny
584
0
Hi, I'm having trouble with the following question.

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

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

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 [itex]curl\mathop F\limits^ \to = \nabla \times \mathop F\limits^ \to [/itex]

[tex]
= \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)
[/tex]

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?
 
Last edited:
Physics news on Phys.org
  • #2
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 [tex] = \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)[/tex]

Good grief! You've correctly analyzed a problem in Green's theorem and messed up the "power law"?? The derivative of xn is nxn-1, not (n-1)xn-1 as you have!
 
  • #3
Heh, I don't know how I managed to miss my error with the power rule. Thanks for the help HallsofIvy.
 

FAQ: Line integral and vector fields

1. What is a line integral?

A line integral is a type of integral that is used to calculate the total value of a scalar or vector field along a specific curve or path. It takes into account the direction and magnitude of the field at each point along the curve.

2. What is a vector field?

A vector field is a mathematical function that assigns a vector (such as velocity or force) to each point in a given area or space. It can be represented graphically by arrows whose direction and length correspond to the direction and magnitude of the vector at each point.

3. How is a line integral calculated?

A line integral can be calculated using the formula ∫F(x,y) · dr = ∫F(x(t),y(t)) · r'(t) dt, where F is the vector field, r is the curve, and t is the parameter of the curve. This formula takes into account the direction and magnitude of the field at each point along the curve.

4. What is the significance of line integrals in physics?

Line integrals are significant in physics as they allow us to calculate the work done by a force along a specific path, which is important in fields such as mechanics and electromagnetism. They also help in understanding the flow of a vector field and the circulation of a fluid.

5. How are line integrals and path independence related?

Line integrals are said to be path independent if their value is the same regardless of the path taken. This means that the starting and ending points of the curve do not affect the result. A vector field is conservative if its line integral is path independent, and this has important implications in solving certain types of problems in physics and engineering.

Similar threads

Back
Top