In this problem, we are asked to compute the line integral

\[

\int_C \frac{-y, dx + x, dy}{x^2 + y^2}

\]

where \(C\) is the line segment from \((1,0)\) to \((0,1)\). It turns out that this integral is particularly easy to compute in polar coordinates (although the work that I did in class probably made it seem hopelessly complicated!).

In polar coorinates, we have \(x = r \cos \theta\) and \(y = r \sin \theta\). To get \(dx\) and \(dy\) in terms of \(r\), \(\theta\) and \(d\theta\) we have to do a little work differentiating. Since we are integrating over the curve \(C\), we will eventually have \(r = r(\theta)\), a function of theta. We compute

\[

dx = d(r \cos \theta) = \left(\frac{dr}{d\theta} \cos \theta – r \sin \theta \right), d\theta

\]

and similarly

\[

dy = d(r \sin \theta) = \left(\frac{dr}{d\theta} sin \theta + r \cos \theta \right), d\theta.

\]

Using these formulas, the numerator of the integrand simplifies immensely:

\[

– y, dx + x, dy = r^2 , d\theta.

\]

Therefore, the integral becomes (in polar coordinates)

\[

\int_C \frac{-y, dx + x, dy}{x^2 + y^2} = \int_C , d\theta.

\]

The curve \(C\) can be written in polar coordinates as

\[

r(\theta) = \frac{1}{\cos \theta + \sin \theta}

\]

where \(0 \leq \theta \leq \pi / 2\). To see this, notice the line satisfies \(y = 1 – x\), or equivalently \(x + y = 1\). In polar coordinates, this equation is \(r \cos \theta + r \sin \theta = 1\), and solving for \(r\) gives the equation for \(C\). Therefore, we can compute the integral

\[

\int_C, d\theta = \int_0^{\pi / 2} , d\theta = \frac{\pi}{2}.

\]

So despite the imposing setup to the problem, the final integral turns out to be very simple.