Question

Let \( R \) be the planar region bounded by the lines \( x = 0 \), \( y = 0 \) and the curve \( x^2 + y^2 = 4 \) in the first quadrant. Let \( C \) be the boundary of \( R \), oriented counter clockwise. Then, the value of:

\[ \oint_C x(1 - y) \, dx + (x^2 - y^2) \, dy \] is equal to:

• A. 0
• B. 2
• C. 4
• D. 8

Hint:

Recognizing when to apply Green's, Stokes', or the Divergence theorem is key. If you have a line integral in 2D over a closed loop, Green's Theorem is almost always the best approach, especially if the resulting double integral is simpler. Converting to polar coordinates for circular regions is a standard and powerful technique.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem asks for the evaluation of a line integral over a closed path \(C\). The path encloses a region \(R\), and the functions involved are polynomials. This is a classic application of Green's Theorem, which relates a line integral around a simple closed curve \(C\) to a double integral over the plane region \(R\) it encloses.

Step 2: Key Formula or Approach:
Green's Theorem states that if \(P(x,y)\) and \(Q(x,y)\) have continuous partial derivatives in a region \(R\) bounded by a piecewise smooth, simple closed curve \(C\) oriented counterclockwise, then: \[ \oint_C P dx + Q dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] In this problem, we have: \( P(x,y) = x(1-y) = x - xy \)
\( Q(x,y) = x^2 - y^2 \)

Step 3: Detailed Explanation:
First, we calculate the partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x - xy) = -x \] Now, we apply Green's Theorem: \[ \oint_C P dx + Q dy = \iint_R (2x - (-x)) dA = \iint_R 3x \, dA \] The region \(R\) is a quarter-circle of radius 2 in the first quadrant. It is described by \(x^2+y^2 \le 4\), \(x \ge 0\), \(y \ge 0\). Due to the circular nature of the region, it's best to switch to polar coordinates.
The transformation is: \(x = r \cos\theta\), \(y = r \sin\theta\), and \(dA = r \, dr \, d\theta\).
The limits of integration for \(R\) in polar coordinates are: \( 0 \le r \le 2 \)
\( 0 \le \theta \le \frac{\pi}{2} \)
Now, we set up and evaluate the double integral: \[ \iint_R 3x \, dA = \int_{0}^{\pi/2} \int_{0}^{2} 3(r \cos\theta) \cdot r \, dr \, d\theta \] \[ = \int_{0}^{\pi/2} \int_{0}^{2} 3r^2 \cos\theta \, dr \, d\theta \] First, integrate with respect to \(r\): \[ \int_{0}^{2} 3r^2 \cos\theta \, dr = \cos\theta \left[ r^3 \right]_{0}^{2} = \cos\theta (2^3 - 0^3) = 8 \cos\theta \] Now, integrate the result with respect to \(\theta\): \[ \int_{0}^{\pi/2} 8 \cos\theta \, d\theta = 8 \left[ \sin\theta \right]_{0}^{\pi/2} = 8 \left( \sin\frac{\pi}{2} - \sin 0 \right) = 8(1 - 0) = 8 \]
Step 4: Final Answer:
The value of the line integral is 8.