Question

If \(\vec r\) is the position vector of any point on a closed surface \(S\) that encloses the volume \(V\), then \(\iint_S \vec r\cdot d\vec s\) is equal to

• A. \(V\)
• B. \(2V\)
• C. \(3V\)
• D. \(\dfrac{V}{2}\)

Hint:

For \(\vec r=x\hat i+y\hat j+z\hat k\), \(\nabla\cdot\vec r=3\).

Answer 1

The Correct Option is C

Concept:
By Gauss divergence theorem, \[ \iint_S \vec F\cdot d\vec s=\iiint_V \nabla\cdot \vec F\,dV \] Here, \[ \vec F=\vec r=x\hat i+y\hat j+z\hat k \]

Step 1: Find divergence of \(\vec r\). \[ \nabla\cdot\vec r = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} \] \[ =1+1+1 \] \[ =3 \]

Step 2: Apply divergence theorem. \[ \iint_S \vec r\cdot d\vec s = \iiint_V 3\,dV \] \[ =3\iiint_V dV \] \[ =3V \]

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