Problem 17.2 #31

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.

Will Rosenbaum

Tel Aviv

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