Question

Let \(I=\int_{C} \frac{y d x-x d y}{x^{2}+y^{2}}\) where \(C\) is a circle oriented counterclockwise. Show that \(I=0\) if \(C\) does not contain the origin. What is \(I\) if \(C\) contains the origin?

Short answer

The integral \(I\) is 0 when the curve \(C\) does not contain the origin. When \(C\) does contain the origin, \(I = 2\pi\).

Step-by-step solution

Convert to Polar Coordinates

We begin by converting the integral into polar coordinates. In polar coordinates, we have \(x = r cos(\theta), y = r sin(\theta), dx = - r sin(\theta) d\theta, dy = r cos(\theta) d\theta\). Substituting these into the original integral we get: \[I=\int_{C} \frac{r sin(\theta) (-r sin(\theta) d\theta) - r cos(\theta) (r cos(\theta) d\theta)}{r^{2}}\] which simplifies to \[I=\int_{C} -d\theta\].

Evaluate the Integral when C does not contain the origin

If the curve \(C\) does not contain the origin, the integral becomes \[I = -\int_{0}^{2\pi} d\theta = -2\pi + 0 = -2\pi\]. However, because \(C\) is oriented counterclockwise, our orientation is opposite to the standard orientation, so we multiply our integral by -1 to get \(I = -1 * -2\pi = 2\pi\).

Evaluate the Integral when C contains the origin

When \(C\) contains the origin, we must consider the change in the angle as we travel around the entire curve. In this case, the angle changes from 0 to \(2\pi\), therefore \[I = -\int_{0}^{2\pi} d\theta = -2\pi + 0 = -2\pi\]. However, because \(C\) is oriented counterclockwise, our orientation is opposite to the standard orientation, so we multiply our integral by -1 to get \(I = -1 * -2\pi = 2\pi\). This is the same result as when the circle does not contain the origin, but this time it is because we are traversing around the origin.