Question

Suppose \(C\) is the closed curve defined as the circle \(x^2+y^2=1\) with \(C\) oriented anti-clockwise. The value of the line integral \[ \oint_C (x\,dy-y\,dx) \] is equal to ____.

• A. \(0\)
• B. \(\pi\)
• C. \(2\pi\)
• D. \(\dfrac{\pi}{2}\)

Hint:

A standard result: \[ \boxed{ \oint_C(x\,dy-y\,dx) = 2\times(\text{Area enclosed}) } \] For the unit circle, \[ \boxed{ 2\times\pi=2\pi. } \]

Answer 1

The Correct Option is C

Concept: Using Green's theorem, \[ \oint_C(P\,dx+Q\,dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)dA. \] Here, \[ P=-y,\qquad Q=x. \]

Step 1: Compute the partial derivatives. \[ \frac{\partial Q}{\partial x}=1, \] \[ \frac{\partial P}{\partial y}=-1. \] Therefore, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1-(-1) = 2. \]

Step 2: Evaluate the double integral. The enclosed region is the unit circle. Hence, \[ \text{Area} =\pi. \] Therefore, \[ \oint_C(x\,dy-y\,dx) = 2\iint_R dA = 2(\pi) = 2\pi. \] Thus, \[ \boxed{ \oint_C(x\,dy-y\,dx)=2\pi. } \] Therefore, the correct option is \[ \boxed{(C)\;2\pi.} \]