Question

Consider a volume V enclosed by a closed surface S having unit surface normal \(\hat{n}\). For \(\mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}\), the value of the surface integral \(\frac{1}{9} \oint_{S} \mathbf{r} \cdot \hat{n} \,dS\) is

• A. \(V\)
• B. \(3V\)
• C. \(\frac{V}{3}\)
• D. \(\frac{V}{9}\)

Hint:

For vector calculus problems, it's extremely useful to remember standard results for the position vector \(\mathbf{r}\). The divergence \(\nabla \cdot \mathbf{r} = 3\) and the curl \(\nabla \times \mathbf{r} = 0\) are frequently used and can save a lot of time in calculations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires the evaluation of a surface integral over a closed surface. The Gauss's Divergence Theorem is the most appropriate tool for this, as it relates a closed surface integral to a volume integral over the volume enclosed by the surface.
Step 2: Key Formula or Approach:
The Gauss's Divergence Theorem states that for a continuously differentiable vector field \(\mathbf{F}\), the outward flux through a closed surface \(S\) is equal to the volume integral of the divergence of \(\mathbf{F}\) over the volume \(V\) enclosed by the surface.
\[ \oint_{S} \mathbf{F} \cdot \hat{n} \,dS = \iiint_{V} (\nabla \cdot \mathbf{F}) \,dV \] Step 3: Detailed Explanation:
In this problem, the vector field is given by the position vector \(\mathbf{F} = \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k}\).
First, we need to calculate the divergence of \(\mathbf{r}\), which is \(\nabla \cdot \mathbf{r}\).
The divergence operator \(\nabla \cdot\) is defined as \(\left( \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right) \cdot\).
\[ \nabla \cdot \mathbf{r} = \left( \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) \] \[ \nabla \cdot \mathbf{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} \] \[ \nabla \cdot \mathbf{r} = 1 + 1 + 1 = 3 \] Now, we apply the Divergence Theorem:
\[ \oint_{S} \mathbf{r} \cdot \hat{n} \,dS = \iiint_{V} (\nabla \cdot \mathbf{r}) \,dV = \iiint_{V} 3 \,dV \] Since 3 is a constant, we can take it out of the integral:
\[ \oint_{S} \mathbf{r} \cdot \hat{n} \,dS = 3 \iiint_{V} dV \] The integral \(\iiint_{V} dV\) simply represents the total volume \(V\) enclosed by the surface \(S\).
\[ \oint_{S} \mathbf{r} \cdot \hat{n} \,dS = 3V \] The question asks for the value of \(\frac{1}{9} \oint_{S} \mathbf{r} \cdot \hat{n} \,dS\).
\[ \frac{1}{9} \oint_{S} \mathbf{r} \cdot \hat{n} \,dS = \frac{1}{9} (3V) = \frac{V}{3} \] Step 4: Final Answer:
The value of the given surface integral is \(\frac{V}{3}\). Therefore, option (C) is the correct answer.