Question

Consider the subset \( S = \{ (x, y) : x^2 + y^2>0 \} \) of \(\mathbb{R}^2.\) Let \[ P(x,y) = \frac{y}{x^2 + y^2}, \quad Q(x,y) = \frac{-x}{x^2 + y^2}. \] For \((x, y) \in S.\) If \(C\) denotes the unit circle traversed in the counter-clockwise direction, then the value of \[ \frac{1}{\pi} \int_C (P\,dx + Q\,dy) \] is _________.

Hint:

Vector fields of the form \(P = \frac{y}{x^2+y^2}, Q = -\frac{x}{x^2+y^2}\) correspond to circular fields. The integral over a closed loop gives a multiple of \(2\pi\), often linked to circulation.

Answer 1

Correct Answer: -2

Step 1: Parameterize the unit circle.
Let \(x = \cos t,\ y = \sin t,\ 0 \le t \le 2\pi.\) Then \(dx = -\sin t\,dt,\ dy = \cos t\,dt.\)
Step 2: Substitute in the integral.
\[ P = \frac{\sin t}{1}, \quad Q = \frac{-\cos t}{1}. \] Hence, \[ P\,dx + Q\,dy = (\sin t)(-\sin t\,dt) + (-\cos t)(\cos t\,dt) = -(\sin^2 t + \cos^2 t)\,dt = -dt. \]
Step 3: Integrate around the circle.
\[ \int_C (P\,dx + Q\,dy) = \int_0^{2\pi} (-dt) = -2\pi. \]
Step 4: Compute final expression.
\[ \frac{1}{\pi} \int_C (P\,dx + Q\,dy) = \frac{-2\pi}{\pi} = -2. \]
Step 5: Conclusion.
The required value is \(\boxed{-2}.\)