Jtdylktuy

I am to consider two paths traveled below:

With the following vector field:

$$vec F = frac{-y hat x + xhat y}{x^2+y^2}$$

And I am to consider the work done going from $(-1,0)$ to $(1,0)$ with the circular and straight path. The work integral is:

$$int_C vec F cdot dvec r$$

The circular path I find is relatively straightforward (as this vector field looks a lot nicer in polar), however computing the line path is something I have difficulties with. My work integral looks like:

$$W = int_{C} frac{-y}{x^2+y^2}dx + int_{C} frac{x}{x^2+y^2}dy$$

Where $C_1$ and $C_2$ represent each line of the path for $y=x +1$ and $y=-x+1$.

For the $dx$ integral, I need to work out how $y$ depends on $x$ for each line. I've called for the parametrization $y=x+1$ and $y=x-1$ for the appropriate line.

$$ int_{C} frac{-y}{x^2+y^2}dx = - int_{-1}^0 frac{x+1}{x^2 + (x+1)^2} + int_0^1 frac{-x+1}{x^2 + (-x+1)^2}$$

Apart from this looking quite hard to integrate, online calculators put each at$-pi/4$. However, the $y$ integral is something that's causing me difficulty. It goes from $0$ to $1$ to $0$, so I'll need to split it up into two integrals, but what parametrization would I use? Would it be $y=x+1 implies x = y -1$ for one integral and $y = x+1$ for the other? If so, how could I justify this other than a guess? If so, integrating this seemed to produce a messy answer that doesn't seem to look right, but that doesn't mean it's wrong. Is there a more efficient way to go about this?

Along the path:
$x+y = 1$

let $x = 1-t, y = t$

$x' = -1, y' = 1$

$int_0^1 frac {-y x' + x y'}{x^2 + y^2} dt \
int_0^1 frac {1}{t^2 -2t + 1 + t^2} dt \
int_0^1 frac {1}{2t^2 - 2t + 1} dt \
int_0^1 frac {1}{2(t^2 - t + frac 14) + frac 12} dt \
int_0^1 frac {1}{2(t - frac 12)^2 + frac 12} dt $

We need a substitution.

We might have avoided it had we chosen a different peramiterization at the beginning... oh well.

$t - frac 12 =frac {1}{2}tantheta\
dt = frac {1}{2}sec^2theta$

$frac{1}{2}int_{-frac {pi}{4}}^{frac {pi}{4}} frac {sec^2theta}{frac 12 sec^2theta} dtheta \
int_{-frac {pi}{4}}^{frac {pi}{4}} dtheta \
frac {pi}{2}$

And I suspect you will get the identical answer when we look at

$y-x = 1$

In hindsight, $F(x,y)$ is a conservative force.

$nabla (arctan frac yx) = F(x,y)$

Which means that all path integrals with the same endpoints will be the same.