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)
- Who was the founder of Bolshevik Communist party?
- What is the key guide to statecraft in the realist tradition?
- Chronologically arrange the events in the Cold War period.
A. Berlin Wall is constructed
B. Communist China joins the UN
C. Soviet invasion of Czechoslovakia
D. Berlin Blockade
Choose the correct answer from the options given below: Morgenthau's principles of political realism are:
A. Politics is rooted in permanent and unchanging human nature which is basically self centred, self-regarding and self-interested
B. Politics is an autonomous sphere of action and cannot therefore be reduced to morals
C. International Politics is an arena of conflicting self-interests
D. The ethics of international relations is situational ethics which is very different from private morality
Choose the correct answer from the options given below:Who among the following political thinkers consider the anarchical self help system to be a compelling factor for States to maximise their relative power positions?
