The following surface integral is to be evaluated over a sphere for the given steady velocity vector field F = xi + yj + zk defined with respect to a Cartesian coordinate system having i, j and k as unit base vectors. $$\iint\limits_{\text{S}} {\frac{1}{4}\left( {{\text{F}} \cdot {\text{n}}} \right){\text{dA}}}$$ where S is the sphere, x2 + y2 + z2 = 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is A. $$\pi $$ B. $$2\pi $$ C. $$\frac{{3\pi }}{4}$$ D. $$4\pi $$

The correct answer is $\boxed{\frac{{3\pi }}{4}}$.

The surface integral can be evaluated using the divergence theorem, which states that the surface integral of a vector field over a closed surface is equal to the triple integral of the divergence of the vector field over the volume enclosed by the surface. In this case, the vector field is $\text{F} = xi + yj + zk$ and the surface is the sphere $x^2 + y^2 + z^2 = 1$. The divergence of $\text{F}$ is $\text{div}(\text{F}) = 1$. The volume enclosed by the sphere is $\frac{4}{3}\pi$. Therefore, the surface integral is equal to

$$\iint\limits_{\text{S}} {\frac{1}{4}\left( {{\text{F}} \cdot {\text{n}}} \right){\text{dA}}} = \frac{1}{4} \iiint\limits_{\text{V}} {\text{div}(\text{F}){\text{dV}} = \frac{1}{4} \cdot \frac{4}{3}\pi \cdot 1 = \frac{{3\pi }}{4}}$$

The other options are incorrect because they do not take into account the fact that the surface integral is over a sphere.