Question

If \(S\) is any closed surface enclosing a volume \(V\) and \(\vec F=x\hat i+2y\hat j+3z\hat k\), then \(\iint_S \vec F\cdot \hat n\,ds=\)

• A. \(V\)
• B. \(2V\)
• C. \(5V\)
• D. \(6V\)

Hint:

For a closed surface, use Gauss divergence theorem: surface flux equals volume integral of divergence.

Answer 1

The Correct Option is D

Concept:
By Gauss divergence theorem, \[ \iint_S \vec F\cdot \hat n\,ds=\iiint_V \nabla\cdot \vec F\,dV \] where \(S\) is a closed surface enclosing volume \(V\).

Step 1: Find divergence of \(\vec F\). Given, \[ \vec F=x\hat i+2y\hat j+3z\hat k \] So, \[ \nabla\cdot \vec F = \frac{\partial x}{\partial x} + \frac{\partial (2y)}{\partial y} + \frac{\partial (3z)}{\partial z} \] \[ =1+2+3 \] \[ =6 \]

Step 2: Apply divergence theorem. \[ \iint_S \vec F\cdot \hat n\,ds = \iiint_V 6\,dV \] Since \(6\) is constant, \[ =6\iiint_V dV \] But \[ \iiint_V dV=V \] Therefore, \[ \iint_S \vec F\cdot \hat n\,ds=6V \]

Step 3: Final answer. \[ \boxed{6V} \]