The value of the integral

${\oint }_s\vec{r}.\vec{n}ds$

over the closed surface S bounding a volume V, where $\vec{r}=xi+yj+zk$ is the position vector and n̂ is normal to the surface S, is

Integral Evaluation using Divergence Theorem

The problem asks us to find the value of the integral ${\oint }_s\vec{r}.\vec{n}ds$ over a closed surface S bounding a volume V. Here, $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is the position vector, and $\hat{n}$ is the outward unit normal vector to the surface S.

Divergence Theorem Application

This type of integral, involving a vector field dotted with the normal vector over a closed surface, is a classic application of the Divergence Theorem, also known as Gauss's Theorem. The Divergence Theorem provides a relationship between a surface integral (flux) and a volume integral.

The theorem states that for a vector field $\vec{F}$ that is continuously differentiable, the surface integral of $\vec{F}$ over a closed surface S is equal to the volume integral of the divergence of $\vec{F}$ over the volume V enclosed by S.

Mathematically, the Divergence Theorem is given by:

$${\oint }_s\vec{F}.\vec{n}ds={\iiint }_V(\nabla \cdot \vec{F})dV$$

In our given problem, the vector field $\vec{F}$ is the position vector $\vec{r}$:

$$\vec{F}=\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$

Calculating Divergence of the Position Vector

The next step is to calculate the divergence of our vector field $\vec{F}=\vec{r}$. The divergence of a vector field $\vec{F}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}$ is defined as:

$$\nabla \cdot \vec{F}=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$$

For our position vector $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$:

- The x-component $F_x=x$.
- The y-component $F_y=y$.
- The z-component $F_z=z$.

Now, let's find the partial derivatives:

- $\frac{\partial F_x}{\partial x}=\frac{\partial }{\partial x}(x)=1$
- $\frac{\partial F_y}{\partial y}=\frac{\partial }{\partial y}(y)=1$
- $\frac{\partial F_z}{\partial z}=\frac{\partial }{\partial z}(z)=1$

Therefore, the divergence of the position vector $\vec{r}$ is:

$$\nabla \cdot \vec{r}=1+1+1=3$$

Evaluating the Volume Integral

Now we substitute the calculated divergence back into the Divergence Theorem equation:

$${\oint }_s\vec{r}.\vec{n}ds={\iiint }_V(\nabla \cdot \vec{r})dV$$ $${\oint }_s\vec{r}.\vec{n}ds={\iiint }_{V}3dV$$

The constant '3' can be taken out of the integral:

$${\oint }_s\vec{r}.\vec{n}ds=3{\iiint }_{V}dV$$

The term ${\iiint }_{V}dV$ represents the total volume V enclosed by the surface S.

So, the integral simplifies to:

$${\oint }_s\vec{r}.\vec{n}ds=3V$$

Summary of Steps

Here's a quick recap of the steps followed:

1. Identified the given integral as a surface integral over a closed surface, suitable for the Divergence Theorem.
2. Recognized the vector field $\vec{F}$ as the position vector $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$.
3. Calculated the divergence of $\vec{r}$, which was $\nabla \cdot \vec{r}=3$.
4. Applied the Divergence Theorem to convert the surface integral into a volume integral of the divergence.
5. Evaluated the volume integral ${\iiint }_{V}3dV$, which resulted in $3V$.

The value of the integral is $3V$.