# Homework Help: Vector field dot product integration!

1. Nov 14, 2012

### Unart

1. The problem statement, all variables and given/known data
Calculate F=∇V, where V(x,y,z)= xye^z, and computer ∫F"dot"ds, where
A)C is any curve from (1,1,0) to (3,e-1)
B)C is a the boundary of the square 0≤x≤1, 0≤y≤1... oriented counterclockwise.

2. Relevant equations
∫F"dot"ds= ∫F(c(t)"dot"c'(t)

3. The attempt at a solution

So my question is how do I pick a curve from the two dots? And, how is it different from a square?

2. Nov 14, 2012

### LCKurtz

Doesn't the suggestion to use "any curve" suggest that it doesn't matter what curve you use? Is that the case in this problem? What do you know about independence of path and potential functions etc.?

Last edited: Nov 14, 2012
3. Nov 15, 2012

### Unart

I know that the V(Q)-V(P) should equal the integral for F dot ds along any path. And that's about it. How to approach it I'm confused....
Do I just, do the subtraction of products of the function? Or, do I have to invent a curve too....

How would I approach the square when the z variable isn't mentioned.

This is my first time doing this.

4. Nov 15, 2012

### LCKurtz

That is an obvious typo. Does it mean (3,e,-1) or what?

Oriented counterclockwise when viewed from where? Above or below the xy plane? And I suppose z = 0??

First you have to check whether your integral is independent of path and the vector field has a potential function. What is the test for that? And if it does have a potential function $\Phi$ then, yes, the value of the integral is the difference of $\Phi$ at the end points.

5. Nov 19, 2012

### Unart

It does... the function is continuous...

6. Nov 19, 2012

### LCKurtz

Just because a vector field is continuous doesn't mean it has a potential function. There are theorems about gradients and curls and potential functions. What do they tell you for this problem?

[Edit, added]: Have you had the "fundamental theorem of calculus for line integrals"?$$\int_P^Q \nabla \Phi \cdot d\vec R=\, \, ?$$

Last edited: Nov 19, 2012