Question

Let \( f: \mathbb{R}^3 \to \mathbb{R} \) be a twice continuously differentiable scalar field such that \( \text{div}(\nabla f) = 6 \). Let \( S \) be the surface \( x^2 + y^2 + z^2 = 1 \) and \( \hat{n} \) be the unit outward normal to \( S \). Then the value of \[ \iint_S (\nabla f \cdot \hat{n}) \, dS \] is

• A. \( 2 \pi \)
• B. \( 4 \pi \)
• C. \( 6 \pi \)
• D. \( 8 \pi \)

Hint:

To evaluate surface integrals of vector fields, use the Divergence Theorem to convert the surface integral into a volume integral.

Answer 1

The Correct Option is D

We are given that \( \text{div}(\nabla f) = 6 \), which is the Laplacian of \( f \), and the surface is the unit sphere \( x^2 + y^2 + z^2 = 1 \). The surface integral of \( \nabla f \cdot \hat{n} \) is equivalent to the flux of \( \nabla f \) through the surface. By the Divergence Theorem, the flux is equal to the volume integral of \( \text{div}(\nabla f) \), which is constant and equal to 6. The volume of the unit sphere is \( \frac{4}{3} \pi \). Hence, the flux is: \[ \iint_S (\nabla f \cdot \hat{n}) \, dS = \int_V \text{div}(\nabla f) \, dV = 6 \times \frac{4}{3} \pi = 8 \pi. \] Therefore, the value of the surface integral is \( 8 \pi \), corresponding to option (A).