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 \dots\dots.

• A. \( 0 \)
• B. \( \pi \)
• C. \( 2\pi \)
• D. Other value

Hint:

The area of any enclosed two-dimensional region can be calculated using the line integral identity \(\text{Area} = \frac{1}{2} \oint_C (x \, dy - y \, dx)\). Rearranging this relationship yields \(\oint_C (x \, dy - y \, dx) = 2 \times \text{Area}\). For a unit circle, this evaluates to \(2 \times \pi = 2\pi\).

Answer 1

The Correct Option is C

Concept: Green's Theorem in a plane relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). It is formulated as: \[ \oint_C (M \, dx + N \, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx \, dy \] where \( M \) and \( N \) are continuous functions with continuous first-order partial derivatives throughout region \( R \).

Step 1: Identify Functions \( M \) and \( N \)
Rearrange the given line integral expression to align with standard notation: \[ \oint_C (-y \, dx + x \, dy) \] From this arrangement, we identify: \[ M = -y \] \[ N = x \]

Step 2: Calculate Partial Derivatives
Compute the partial derivatives required for the integrand of Green's double integral: \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x) = 1 \] \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(-y) = -1 \] Substitute these values into Green's integrand: \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - (-1) = 2 \]

Step 3: Evaluate the Double Integral over Region \( R \)
Substitute the integrand into the double integral expression: \[ \oint_C (x \, dy - y \, dx) = \iint_R 2 \, dx \, dy = 2 \iint_R dx \, dy \] The term \(\iint_R dx \, dy\) represents the total geometric area of region \( R \). Since \( C \) is the unit circle \( x^2 + y^2 = 1 \), region \( R \) is a unit disk with radius \( r = 1 \). The area of a circle is given by: \[ \text{Area} = \pi r^2 = \pi (1)^2 = \pi \] Substitute this area back into the equation: \[ \oint_C (x \, dy - y \, dx) = 2 \times \pi = 2\pi \] Thus, the value of the line integral is \( 2\pi \), matching Option (C).