To solve this problem, we need to evaluate the line integral \(\oint_C x(1-y)dx + (x^2-y^2)dy\) over the closed curve \(C\). The region \(R\) is bounded by the lines \(x = 0\), \(y = 0\), and the curve \(x^2 + y^2 = 4\) in the first quadrant. This means \(R\) is the quarter-circle of radius 2, centered at the origin, in the first quadrant.
The boundary \(C\) therefore consists of three parts:
- The arc of the circle from \((2, 0)\) to \((0, 2)\), expressed parametrically as \((x, y) = (2\cos t, 2\sin t)\) for \(t \in \left[0, \frac{\pi}{2}\right]\).
- The vertical line segment along the \(y\)-axis from \((0, 2)\) to \((0, 0)\).
- The horizontal line segment along the \(x\)-axis from \((0, 0)\) to \((2, 0)\).
Since the integral \(\oint_C P\,dx + Q\,dy\) is given, where \(P = x(1-y)\) and \(Q = x^2 - y^2\), we can use 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:
$$\oint_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dx \, dy.$$Calculating the partial derivatives:
- \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x\)
- \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x(1-y)) = -x\)
Substituting these into Green’s Theorem gives:
$$\iint_R (2x - (-x)) \, dx \, dy = \iint_R 3x \, dx \, dy.$$To evaluate this double integral, convert to polar coordinates since \(R\) is a quarter-circle:
- The transformation is \(x = r\cos\theta\) and \(y = r\sin\theta\), with \(r \in [0, 2]\) and \(\theta \in \left[0, \frac{\pi}{2}\right]\).
- The Jacobian of the transformation is \(r\), so \(dx \, dy = r \, dr \, d\theta\).
Substituting, we get:
$$\iint_R 3x \, dx \, dy = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} 3(r\cos \theta) \cdot r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} 3r^2 \cos\theta \, dr \, d\theta.$$Evaluating the inner integral with respect to \(r\):
$$\int_{0}^{2} 3r^2 \, dr = \left[r^3\right]_{0}^{2} = 8.$$So the integral becomes:
$$\int_{0}^{\frac{\pi}{2}} 8\cos\theta \, d\theta.$$Evaluating the integral with respect to \(\theta\):
$$\int_{0}^{\frac{\pi}{2}} 8\cos\theta \, d\theta = [8\sin\theta]_{0}^{\frac{\pi}{2}} = 8.$$Hence, the value of the line integral \(\oint_C x(1-y)dx + (x^2-y^2)dy\) is:
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Important Questions from Mixed Topic (CUET PG)
- Kalpsutra, the illustrated canonical text is from:-
- The Harappan city almost exclusively devoted to craft production was-:
- Mohandas Karamchand Gandhi launched quit India movement after the failure of:-
- The "Objectives Resolution" was introduced in constituent assembly by:-
- Who among the following was not a member of the constituent assembly:-
