# Finding Line Integral of Vector Field

1. Sep 19, 2010

### xxbigelxx

1. The problem statement, all variables and given/known data
You are given a vector field A= kx2 x.
a. First, calculate the line integral of A from x=-2 to x=2 along the x axis.
b. Next, calculate the line integral of A between the same 2 points, but use
a semicircular path with a center at the origin. Recall that in cylindrical
coordinates, dl= ds s + sdφ φ + dz z.
c. Are the 2 results the same? Explain.

2. Relevant equations

3. The attempt at a solution

I believe I did the first part correctly and I got (16/3)k. I am getting confused on the second part though. My value for dl ended up being 2d(phi)phi. This is as far as I could get though.

2. Sep 20, 2010

### vela

Staff Emeritus
So the integrand is equal to

$$\vec{A}\cdot d\vec{l} = (kx^2\,\hat{x})\cdot(s\,d\phi\,\hat{\phi}) = kx^2s\,d\phi\,(\hat{x}\cdot\hat{\phi})$$

What is $$\hat{x}\cdot\hat{\phi}$$ equal to?

Last edited: Sep 20, 2010
3. Sep 20, 2010

### xxbigelxx

I'm not sure. Should I change the 'x' into cylindrical form? That would make x=rcos(theta).

4. Sep 20, 2010

### HallsofIvy

Is x the "position vector" <x, y>? On the x-axis, of course, that is <x, 0> so the vector field, restricted to the x-axis is <kx^3, 0>. The integral, then, from x=-2 to 2 is
$$\int_{-2}^2 kx^3dx= \frac{k}4}x^4\right|_{-2}^2= 0$$

Or is x the unit vector in the x direction? (I would use $\vec{i}$ for that. If so, then the vector field is just <kx^2, 0> and the integral is
$$\int_{-2}^2 kx^2 dx= \frac{k}{3}x^3\right|_{-2}^2= \frac{16k}{3}$$.

I would NOT change to cylindrical form. In Cartesian coordinates, the upper semi circle from -2 to 2 can be written in parametric form as x= 2cos(t), y= 2sin(t), t going from $-\pi$ to 0. $d\vec{s}= (-2sin(t)\vec{i}+ 2cos(t)\vec{j}) dt. With x= <x, y>, [itex]kx^2$x is $<x^3, x^2y>$ and on the upper semi-circle that is $8sin^3(t)\vec{i}+ 8sin(t)cos^2(t)\vec{j}$

The integral would be
$$16\int_{-\pi}^0 -sin^4(t)+ sin(t)cos^3(t) dt$$
For the first,$sin^4(t)$, use the trig identity $sin^2(t)= (1/2)(1- cos(2t)$. For the second, $sin(t)cos^3(t)$, use the substitution u= cos(t), du= -sin(t)dt. When $t= -\pi$, u= -1 and when t= 0, u= 1.

5. Sep 20, 2010

### vela

Staff Emeritus
If you're referring to the variable $x$ and not the unit vector $\hat{x}$, then yes, you need to express it in terms of the cylindrical variables in the integrand. You're going to be integrating with respect to $\phi$, so any quantity that varies with $\phi$ has to be expressed in terms of it.

In your textbook or in your notes, you have probably been told what $$\hat{\phi}$$ is equal to. Use it to evaluate the dot product.

6. Sep 20, 2010

### xxbigelxx

HallsofIvy: I see what you are saying, but I think the teacher wants us to convert to cylindrical coordinates. I think that's why he put 'dl' in terms of cylindrical coordinates in the problem.
Vela: I don't have anything in my notes, but I found this online...Phi-hat= -sin(phi)(x-hat) + cos(phi)(y-hat). Is that the appropriate form?
If so, if I dot that value with x-hat, I just get -sin(phi)?

7. Sep 20, 2010

### vela

Staff Emeritus
Yes, that's exactly what you want. $$\hat{\phi}$$ points in the direction a point would move if you increase φ infinitesimally. If you try evaluate $$\hat{\phi}$$ for a few values of φ, you should see that expression is correct.

8. Sep 20, 2010

### xxbigelxx

Ok so I plugged x=Rcos(phi) and Phi-hat= -sin(phi)(x-hat) + cos(phi)(y-hat) into what I had. Shouldn't I have a dPhi somewhere though?

9. Sep 20, 2010

### vela

Staff Emeritus
Yes. What did you do with it?

10. Sep 20, 2010

### xxbigelxx

Ohhh I found it. Ok so here is my work. I think I got it. Also for part C, I just stated that they were supposed to be equal since line integrals are path independent.

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11. Sep 20, 2010

### vela

Staff Emeritus
I couldn't read part of it, but the set-up looked fine. For part (c), remember that line integrals aren't always path-independent. You should show that this vector field satisfies the condition required for path independence.

12. Sep 20, 2010

### xxbigelxx

Well I ended up getting 16k/3 for both part A and B, so I hope it's correct. And for part C, I know that independence requires a line integral of a vector field to be equal to the gradient of a scalar field.

13. Sep 20, 2010

### vela

Staff Emeritus
I didn't mean to cast any doubt on your work. I just didn't want to imply that I had checked all of it carefully. I think you did everything correctly.

Can you guess a scalar field Φ that satisfies $\vec{A} = \nabla \phi$?

14. Sep 20, 2010

### xxbigelxx

Oh ok.

And I honestly am not sure how you even got that to start with. I understand the general format, but I don't see why it's the gradient of Phi. Phi is the angle that the particle travels, but I am still confused on the relation that allowed you to write gradientPhi

15. Sep 20, 2010

### vela

Staff Emeritus
Oops, I'm sorry. I should have used a different symbol. I meant just some scalar function (phi is common notation for such a function). It has nothing to do with the angle.

16. Sep 20, 2010

### xxbigelxx

Ohhh ok. I understand it now, but I still don't know what the function would be. Would it be
(kx^3)/3?

17. Sep 20, 2010

### vela

Staff Emeritus
Yes, that works.

18. Sep 20, 2010

### xxbigelxx

So should I say something like this.....

As long as the relationship of A=gradient((kx^3)/3) holds, then the line integral is path independent.

19. Sep 20, 2010

### vela

Staff Emeritus
It would be better to say, "Because the vector field A can be expressed as a gradient of the scalar function kx3/3, blah blah blah."

20. Sep 20, 2010

### xxbigelxx

Ok I see what you are saying. I wrote this....

Yes because the vector filed A can be expressed as a gradient of the scalar function kx^3/3. We know that if the line integral of a vector field is equal to the gradient of a scalar field, the line integral is path independent.