Question

Apply Gauss’s Divergence theorem to evaluate taken over the

sphere being the direction cosines of the external normal to the

sphere.

Answer 1

Here the given question is not complete and it is not from the related subject. So, I am answering the approach of the above question.

Gauss Divergence Theorem:

The surface integral of a vector A over the closed surface = Volume integral of the divergence of a vector field A over the volume enclosed by closed surface.

$\Rightarrow \oiint_s \overrightarrow{A}.\overrightarrow{dS}=\iiint_v(\overrightarrow{{\nabla}}.\overrightarrow{A})dV$

Let there is a surface S, which encloses a surface A. Let A be the vector field in the given region.

Let the volume of the small part of the sphere $\Delta V_j$ which is bounded by a surface $S_j$

$\Rightarrow \oiint_s \overrightarrow{A}.\overrightarrow{dS}$

Now, we will integrate the volume,

$\Rightarrow \Sigma \oiint_{sj} \overrightarrow{A}.\overrightarrow{dS_j}=\oiint_s \overrightarrow{A}.\overrightarrow{dS}...(i)$

Now, in the above equation, multiply and divide by the $\Delta V_i$

$\Rightarrow \oiint_s \overrightarrow{A}.\overrightarrow{dS}=\Sigma{\frac{1}{\Delta{V_i}}}(\oiint\overrightarrow{A}\overrightarrow{dS_j})\Delta{V_i}$

Let volume is divide into the infinite elementary volume,

$\Rightarrow \oiint_s \overrightarrow{A}.\overrightarrow{dS}=\lim_{\Delta v_i\rightarrow 0}\Sigma{\frac{1}{\Delta{V_i}}}(\iint\overrightarrow{A}\overrightarrow{dS_j})\Delta{V_i}...(ii)$

$\lim_{\Delta v_i\rightarrow 0}[\Sigma{\frac{1}{\Delta{V_i}}}(\oiint\overrightarrow{A}\overrightarrow{dS_i})]=(\overrightarrow{V}.\overrightarrow{A})$

$\oiint{(\overrightarrow{A}.\overrightarrow{dS})}=\Sigma{(\overrightarrow{\nabla}.\overrightarrow{A})}\Delta{V}_i$

Hence,

$\oiint{(\overrightarrow{A}.\overrightarrow{dS})}=\iiint_v(\overrightarrow{\nabla}.\overrightarrow{A})dV$