Given a function$\varphi =\frac{1}{2}({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2)$ in three-dimensional Cartesian space, the value of the surface integral ∯S n̂ . ∇ϕ dS where S is the surface of a sphere of unit radius and n̂ is the outward unit normal vector on S, is

The correct option is 4π.

To find the value of the surface integral, we can use two different methods: the Gauss Divergence Theorem (which is highly efficient) and direct integration. Below is the step-by-step explanation for both methods.

Method 1: Using the Gauss Divergence Theorem

The Gauss Divergence Theorem states that the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence of that vector field over the region enclosed by the surface:
${∯}_S\hat{n}\cdot \mathbf{F}dS={\iiint }_V\nabla \cdot \mathbf{F}dV$
Here, the vector field is $\mathbf{F}=\nabla \varphi$, where:
$\varphi =\frac{1}{2}(x^2+y^2+z^2)$

First, we calculate the gradient of $\varphi$:
$\nabla \varphi =\frac{\partial \varphi }{\partial x}\hat{i}+\frac{\partial \varphi }{\partial y}\hat{j}+\frac{\partial \varphi }{\partial z}\hat{k}$
Taking the partial derivatives:
$\frac{\partial \varphi }{\partial x}=\frac{1}{2}(2x)=x$
Similarly, $\frac{\partial \varphi }{\partial y}=y$ and $\frac{\partial \varphi }{\partial z}=z$.
Therefore, the gradient is:
$\nabla \varphi =x\hat{i}+y\hat{j}+z\hat{k}$

Next, we calculate the divergence of this vector field, which is $\nabla \cdot (\nabla \varphi )={\nabla }^2\varphi$:
$\nabla \cdot (\nabla \varphi )=\frac{\partial }{\partial x}(x)+\frac{\partial }{\partial y}(y)+\frac{\partial }{\partial z}(z)=1+1+1=3$

Now, we substitute this back into the volume integral:
${∯}_S\hat{n}\cdot \nabla \varphi dS={\iiint }_{V}3dV=3{\iiint }_{V}dV$
The volume integral represents the volume of the sphere of unit radius (r = 1). The formula for the volume of a sphere is:
$V=\frac{4}{3}\pi r^3$
For unit radius (r = 1):
$V=\frac{4}{3}\pi (1{)}^3=\frac{4}{3}\pi$
Multiplying by the constant value of the divergence:
$3\times \frac{4}{3}\pi =4\pi$

Method 2: Direct Surface Integration

Alternatively, we can evaluate the surface integral directly over the unit sphere. The outward unit normal vector on the surface of a sphere centered at the origin is given by:
$\hat{n}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{r}$
Since the sphere has unit radius, we have r = 1, which means:
$\hat{n}=x\hat{i}+y\hat{j}+z\hat{k}$

Now, compute the dot product $\hat{n}\cdot \nabla \varphi$:
$\hat{n}\cdot \nabla \varphi =(x\hat{i}+y\hat{j}+z\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=x^2+y^2+z^2$
On the surface S of the unit sphere, we have the constraint equation:
$x^2+y^2+z^2=1$
Thus, the dot product simplifies to exactly 1 at all points on the surface S.

Substitute this back into the surface integral:
${∯}_S\hat{n}\cdot \nabla \varphi dS={∯}_{S}1dS$
This integral is simply the total surface area of the unit sphere. Using the surface area formula $A=4\pi r^2$ with r = 1:
$A=4\pi (1{)}^2=4\pi$

Both methods yield the exact same value of 4π.