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I'm trying to solve the following problem:

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).

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.

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    1. Are you sure that the work along all line segments equal? 2. Have you tried to find the work only on the given line segment $r=(t,-t+1)$? – 2017-01-22
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    1 - Exactly I'm not sure about that... 2 - Yes, I tried by definition but I end up with a quite hard integral... – 2017-01-22

1 Answers 1

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The line segment may be parametrized by $r (t)=(t,-t+1)$ and hence the wok done is given by
$$W=\int_0^1F(r(t)).r'(t) dt\\ =\int_0^1\left(\frac{-t+1}{2t^2-2t+1}, \frac{-t}{2t^2-2t+1} \right).(1,-1)dt\\ =\int_0^1\left(\frac{1}{2t^2-2t+1} \right)dt\\ =\frac 12\int_0^1\left(\frac{1}{(t-\frac12)^2+\frac14} \right)dt$$ Now your turn.