Question

Let S be a closed surface for which \( \oiint_S \vec{r} \cdot \hat{n} \, dS = 1 \), then the volume enclosed by the surface is ____ .

• A. 1
• B. 1/3
• C. 2/3
• D. 3

Hint:

Divergence Theorem. \( \oiint_S \vec{F \cdot \hat{n \, dS = \iiint_V (\nabla \cdot \vec{F) \, dV \). Remember \(\nabla \cdot \vec{r = 3\), where \(\vec{r\) is the position vector.

Answer 1

The Correct Option is B

We are given the surface integral \( \oiint_S \vec{r} \cdot \hat{n} \, dS = 1 \), where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\) is the position vector and \(\hat{n}\) is the outward unit normal vector to the closed surface S
We can apply the Divergence Theorem (Gauss's Theorem), which relates a surface integral of a vector field over a closed surface to the volume integral of the divergence of the field over the volume V enclosed by the surface: $$ \oiint_S \vec{F} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \vec{F}) \, dV $$ In this case, the vector field is \(\vec{F} = \vec{r}\)
We need to find the divergence of \(\vec{r}\): $$ \nabla \cdot \vec{r} = \left( \frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k} \right) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) $$ $$ = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 $$ Applying the Divergence Theorem: $$ \oiint_S \vec{r} \cdot \hat{n} \, dS = \iiint_V (\nabla \cdot \vec{r}) \, dV = \iiint_V 3 \, dV $$ $$ = 3 \iiint_V dV $$ The integral \( \iiint_V dV \) represents the total volume V enclosed by the surface S
So, \( \oiint_S \vec{r} \cdot \hat{n} \, dS = 3V \)
We are given that \( \oiint_S \vec{r} \cdot \hat{n} \, dS = 1 \)
Therefore, \( 3V = 1 \), which implies \( V = 1/3 \)
The volume enclosed by the surface is 1/(3)