Question

The flux of the vector field \[ \mathbf{F} = \left( \frac{2\pi x + 2x^2 y^2}{\pi} \right) \hat{i} + \left( \frac{2\pi x y - 4y}{\pi} \right) \hat{j} \] along the outward normal, across the ellipse \( x^2 + 16y^2 = 4 \) is equal to

• A. \( 4\pi^2 - 2 \)
• B. \( 2\pi^2 - 4 \)
• C. \( \pi^2 - 2 \)
• D. \( 2\pi \)

Hint:

For flux calculations, use the parametrization of the curve and compute the line integral along the path.

Answer 1

The Correct Option is B

Step 1: Flux formula.
The flux of a vector field \( \mathbf{F} = P \hat{i} + Q \hat{j} \) across a curve \( C \) with outward normal vector \( \mathbf{n} \) is given by: \[ \text{Flux} = \oint_C P \, dx + Q \, dy. \]
Step 2: Parametrize the ellipse.
The ellipse \( x^2 + 16y^2 = 4 \) can be parametrized as: \[ x = 2 \cos t, \quad y = \frac{1}{4} \sin t, \quad t \in [0, 2\pi]. \] The outward normal vector on the ellipse is computed by finding the gradient of the equation \( x^2 + 16y^2 = 4 \). The flux across the ellipse is then computed by evaluating the line integral using the parametric equations.
Step 3: Calculation.
By substituting the parametric expressions into the flux formula, and simplifying the resulting integrals, we find that the flux is \( 2\pi^2 - 4 \). Thus, the correct answer is (B).