Step 1: Problem Overview:
The problem requires evaluating a line integral over a closed path \(C\) enclosing a region \(R\), involving polynomial functions. Green's Theorem is applicable here, relating the line integral around a closed curve \(C\) to a double integral over the enclosed region \(R\).
Step 2: Key Formula:
Green's Theorem: 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 case:
\( P(x,y) = x(1-y) = x - xy \)
\( Q(x,y) = x^2 - y^2 \)
Step 3: Solution:
Calculate partial derivatives:
\[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \]
\[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x - xy) = -x \]
Apply Green's Theorem:
\[ \oint_C P dx + Q dy = \iint_R (2x - (-x)) dA = \iint_R 3x \, dA \]
Region \(R\) is a quarter-circle (radius 2) in the first quadrant, described by \(x^2+y^2 \le 4\), \(x \ge 0\), \(y \ge 0\). Convert to polar coordinates: \(x = r \cos\theta\), \(y = r \sin\theta\), and \(dA = r \, dr \, d\theta\). Limits of integration in polar coordinates:
\( 0 \le r \le 2 \)
\( 0 \le \theta \le \frac{\pi}{2} \)
Evaluate the double integral:
\[ \iint_R 3x \, dA = \int_{0}^{\pi/2} \int_{0}^{2} 3(r \cos\theta) \cdot r \, dr \, d\theta \]
\[ = \int_{0}^{\pi/2} \int_{0}^{2} 3r^2 \cos\theta \, dr \, d\theta \]
Integrate with respect to \(r\):
\[ \int_{0}^{2} 3r^2 \cos\theta \, dr = \cos\theta \left[ r^3 \right]_{0}^{2} = \cos\theta (2^3 - 0^3) = 8 \cos\theta \]
Integrate 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: Result:
The line integral evaluates to 8.