Apply Gauss’s Divergence theorem to evaluate taken over the
sphere being the direction cosines of the external normal to the
sphere.
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"
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