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.

Step-by-step Explanation:

We are given the scalar function:

$$\varphi =\frac{1}{2}(x^2+y^2+z^2)$$

We need to evaluate the surface integral:

$${\iint }_S\hat{n}\cdot \nabla \varphi dS$$

where $S$ is the surface of a sphere of unit radius, and $\hat{n}$ is the outward unit normal vector on $S$.

Step 1: Apply Gauss's Divergence Theorem
According to Gauss's Divergence Theorem, the surface integral of a vector field over a closed surface $S$ is equal to the volume integral of the divergence of that vector field over the volume $V$ enclosed by the surface:

$${\iint }_S\nabla \varphi \cdot \hat{n}dS={\iiint }_V\nabla \cdot (\nabla \varphi )dV={\iiint }_V{\nabla }^2\varphi dV$$

where ${\nabla }^2\mathrm{��}$ is the Laplacian of $\varphi$.

Step 2: Calculate the Laplacian of $\varphi$
First, we find the first-order partial derivatives of $\varphi$:

$$\frac{\partial \varphi }{\partial x}=\frac{1}{2}(2x)=x$$

$$\frac{\partial \varphi }{\partial y}=\frac{1}{2}(2y)=y$$

$$\frac{\partial \varphi }{\partial z}=\frac{1}{2}(2z)=z$$

Next, we find the second-order partial derivatives:

$$\frac{{\partial }^2\varphi }{\partial x^2}=1,\frac{{\partial }^2\varphi }{\partial y^2}=1,\frac{{\partial }^2\varphi }{\partial z^2}=1$$

Summing these gives the Laplacian:

$${\nabla }^2\varphi =\frac{{\partial }^2\varphi }{\partial x^2}+\frac{{\partial }^2\varphi }{\partial y^2}+\frac{{\partial }^2\varphi }{\partial z^2}=1+1+1=3$$

Step 3: Evaluate the Volume Integral
Substitute the value of the Laplacian into the volume integral:

$${\iiint }_V{\nabla }^2\varphi dV={\iiint }_{V}3dV=3{\iiint }_{V}dV=3V$$

where $V$ is the volume of the sphere of unit radius ($R=1$).

The volume of a sphere is given by the formula:

$$V=\frac{4}{3}\pi R^3$$

For a unit sphere ($R=1$):

$$V=\frac{4}{3}\pi (1{)}^3=\frac{4}{3}\pi$$

Therefore, the value of the integral is:

$$3V=3\times \frac{4}{3}\pi =4\pi$$