Question

The closed curve shown in the figure is described by \[ r = 1 + \cos\theta, r = \sqrt{x^2 + y^2}; x = r\cos\theta, \; y = r\sin\theta \] The magnitude of the line integral of the vector field \[ F = -y \hat{i} + x \hat{j} \] around the closed curve is ............... (Round off to 2 decimal places). \begin{center} \includegraphics[width=0.5\textwidth]{33.jpeg} \end{center}

Hint:

Green's theorem is very effective to evaluate line integrals of rotational fields: \(\oint F \cdot dl = \iint \text{curl}(F) \cdot \hat{n} \, dA\).

Answer 1

Step 1: Recognize vector field.
\[ F = -y \hat{i} + x \hat{j} \] This corresponds to a rotational field (like circulation around origin).

Step 2: Line integral.
By Green's theorem: \[ \oint F \cdot dl = \iint \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] where \(F = P\hat{i} + Q\hat{j}\). Here: \(P = -y, \; Q = x\). \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \frac{\partial (x)}{\partial x} - \frac{\partial (-y)}{\partial y} = 1 - (-1) = 2 \]

Step 3: Area of region.
Region is cardioid: \(r = 1 + \cos\theta\). Area: \[ A = \frac{1}{2} \int_{0}^{2\pi} r^2 d\theta = \frac{1}{2} \int_{0}^{2\pi} (1 + \cos\theta)^2 d\theta \] \[ = \frac{1}{2} \int_{0}^{2\pi} (1 + 2\cos\theta + \cos^2\theta) d\theta \] \[ = \frac{1}{2} \left[ \int_0^{2\pi} 1 d\theta + 2\int_0^{2\pi}\cos\theta d\theta + \int_0^{2\pi}\frac{1+\cos2\theta}{2} d\theta \right] \] \[ = \frac{1}{2} \left[ 2\pi + 0 + \pi \right] = \frac{3\pi}{2} \]

Step 4: Integral value.
\[ \oint F \cdot dl = 2 \times A = 2 \times \frac{3\pi}{2} = 3\pi \approx 9.42 \]

Final Answer:
\[ \boxed{9.42} \]