Question

Let \( \vec{F}(x,y) = e^x \sin x\, \hat{i + x\, \hat{j} \) for \((x,y)\in \mathbb{R}^2\). If \(C\) is the circle \(x^2 + y^2 = 4\) oriented anticlockwise, then \[ \oint_C \vec{F} \cdot d\vec{R} \] equals:}

• A. \(4\pi\)
• B. \(6\pi\)
• C. \(7\pi\)
• D. \(8\pi\)

Hint:

Use Green’s Theorem on closed, anticlockwise curves to convert a line integral into an easier double integral.

Answer 1

The Correct Option is A

To evaluate the line integral of the vector field \(\vec{F}\) around the circle \(C: x^2 + y^2 = 4\), we use Green’s Theorem, since the curve is closed and traversed anticlockwise.
\[ \oint_C (P\,dx + Q\,dy) = \iint_{R} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA \] Here, \[ P(x,y) = e^x \sin x,\qquad Q(x,y) = x \]
Step 1: Compute the partial derivatives.
\[ \frac{\partial Q}{\partial x} = 1 \] \[ \frac{\partial P}{\partial y} = 0 \quad \text{(since \(P\) has no \(y\)-dependence)} \] Thus, \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 \]
Step 2: Evaluate the double integral over the region \(R\).
The region \(R\) is the disk of radius 2 (because \(x^2 + y^2 = 4\)). So, \[ \iint_R 1\, dA = \text{Area of the disk of radius 2} = \pi (2)^2 = 4\pi \]
Step 3: Conclusion.
Applying Green’s Theorem: \[ \oint_C \vec{F} \cdot d\vec{R} = 4\pi \]
Thus, the line integral equals \(4\pi\).