Question

Given a function \( \varphi = \frac{1}{2} (x^2 + y^2 + z^2) \) in three-dimensional Cartesian space, the value of the surface integral \[ \int \int_S \hat{n} \cdot \nabla \varphi \, dS, \] where \( S \) is the surface of a sphere of unit radius and \( \hat{n} \) is the outward unit normal vector on \( S \), is

• A. \( 4\pi \)
• B. \( 3\pi \)
• C. \( \frac{4\pi}{3} \)
• D. 0

Hint:

Use the divergence theorem for surface integrals of the form \( \int \int_S \hat{n} \cdot \nabla \varphi \, dS \), where \( \nabla \varphi \) is the gradient of the function.

Answer 1

The Correct Option is A

Step 1: Evaluate the Gradient of \( \varphi \).
We first compute the gradient of the function \( \varphi = \frac{1}{2}(x^2 + y^2 + z^2) \): \[ \nabla \varphi = (x, y, z). \] Step 2: Use the Divergence Theorem.
The given surface integral is of the form \( \int \int_S \hat{n} \cdot \nabla \varphi \, dS \), which by the divergence theorem becomes a volume integral of the divergence of \( \nabla \varphi \). The divergence of \( \nabla \varphi \) is: \[ \nabla \cdot \nabla \varphi = 3. \] Step 3: Apply the Divergence Theorem.
The volume of the unit sphere is \( \frac{4\pi}{3} \), so the value of the surface integral is: \[ \int \int_S \hat{n} \cdot \nabla \varphi \, dS = 4\pi. \] Final Answer: \[ \boxed{4\pi} \]