# Multivariable calculus: work in a line segment

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1. Jan 22, 2017

### Granger

1. The problem statement, all variables and given/known data

Compute the work of the vector field $F(x,y)=(\frac{y}{x^2+y^2},\frac{-x}{x^2+y^2})$
in the line segment that goes from (0,1) to (1,0).

2. Relevant equations
3. The attempt at a solution

My attempt (please let me know if there is an easier way to do this)

I applied Green's theorem in the region between the square of vertices (1,0), (0,1), (-1,0), (0,-1), and the circumference centered in the origin with radius 1/2, both clockwise.

Since both lines are clockwise, and because F is field of class $C^1$ then

$\int_C F = \int_S F$ (C circumference and S square).

C is then described by the path $\gamma=(\frac{\cos t}{2},\frac{-\sin t}{2}) t\in]0,2\pi[$

We have $F(\gamma (t)) \gamma ´(t)=1$ so $\int_C F = 2\pi = \int_S F$

Now because we want only the work in the line segment that goes from (0,1) to (1,0) we divide our result by 4 and obtain $\frac{\pi}{2}$

My doubts here is if this is correct, especially the final step... I also wonder if there was an easier way to approach the problem. I first thought of applying the fundamental theorem of calculus but we can't because F is not conservative. Then I tried the definition but we end up with a hard integral to compute. So I ended up with this...

Thanks for the help.

Last edited by a moderator: Jan 22, 2017
2. Jan 22, 2017

### Ray Vickson

Show us the actual integral you get when you apply the definition. I would say that the only correct way to do the problem is by applying the definition; the other things you did have no relation at all to the problem as originally posed.

3. Jan 22, 2017

### pasmith

You can set $x = r(\theta)\cos\theta$, $y = r (\theta)\sin\theta$ and you don't actually need to know that $r(\theta) = (\cos \theta + \sin \theta)^{-1}$ or what $r'(\theta)$ is, because after multiplying it all out and collecting terms it will simplify considerably.

4. Jan 22, 2017

### LCKurtz

You describe the line segment from $(0,1)$ to $(1,0)$. Presumably a straight line. What does that have to do with the circle and square you describe?
Also, you might edit your post and use double  instead of single \$'s to display your tex. Or use $'s for inline. 5. Jan 22, 2017 ### Granger Thanks for all the replies. The integral I obtained by definition was \int \frac{1}{2t^2-2t+1} dt. Any suggestions on how to solve this integral the simplest way? 6. Jan 22, 2017 ### LCKurtz You haven't answered this question: I would complete the square in the denominator. But you need to explain to us what integral you are actually calculating and show your steps so we know what you are talking about. 7. Jan 23, 2017 ### Granger My apologies, the line segment can be described by \gamma (t) = (t,1-t) t from 0 to 1. Then I apply the definition  \int F(\gamma (t)) \gamma ' (t) dt 8. Jan 23, 2017 ### Ray Vickson That definition looks wrong. For example, why could I not take$\vec{\gamma}(u) = (u^2, 1-u^2)$and then have$\int_0^1 \vec{F}(\vec{\gamma}(u)) \cdot \vec{\gamma}'(u) \, du##?

9. Jan 23, 2017

### LCKurtz

OK, that's what I thought you might be doing. That's a dot product in there, which I have inserted.

I think you could Ray. Isn't that just a different parameterization?
Anyway @Granger my suggestion earlier still stands. Complete the square and find an appropriate trig substitution and it will all work out.

10. Jan 23, 2017

### Ray Vickson

Yes, it is a different parametrization, but the question I was aiming at was whether that could alter the outcome. I think I know the answer, but I was hoping the OP would ponder the issue.

11. Jan 24, 2017

### Granger

Looks wrong? What do you mean? Well I know that with other path the the work is still the same (unless the direction was the opposite, in which we would have the opposite sign).

I'm going to try then, thanks.

12. Jan 24, 2017

### Ray Vickson

What I mean is nothing: I was thinking of something else and made a stupid mistake. Your expression is OK, or would be if you could "vectorize" it.