Example of evaluating surface integral of a sphere
Question : Evaluate the surface integral ∬ 𝐹. 𝑛𝑑𝐴 𝑆 for 𝐹 = [𝑥 3 , 𝑦 3 , 𝑧 3 ] for spherical surface 𝑆 = 𝑥^2 + 𝑦^2 + 𝑧^2 = 9
Solution :
To evaluate the surface integral
\[\iint_{\gamma}{\vec{F}.\hat{n}dS\ }\]For the vector field
\[ \vec{F}\ =\left[x^3\ y^3\ z^3\right]\]
for spherical surface
\[S : x^2\ \ +\ y^2\ +\ z^2\ =\ 9\]Using Gauss Divergence theorem
\[\iint_{\gamma}{\vec{F}.\hat{n}dS\ }=\iiint_{V}{\nabla.\vec{F}}dV\]\[=\iiint_{V}{\left(\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}\right).\left(x^3\hat{i}+y^3\hat{j}+z^3\hat{k}\right)}dV\]\[=\iiint_{V}\left(\frac{\partial x^3}{\partial x}+\frac{\partial y^3}{\partial y}+\frac{\partial z^3}{\partial z}\right)dV\]\[=\iiint_{V}\left(3x^2+3y^2+3z^2\right)dV\]
\[=3\iiint_{V}\left(x^2+y^2+z^2\right)dV\]Since for the sphere x^2+y^2+z^2=9
\[=3\iiint_{V}9dV\]
\[=27\iiint_{V}1dV\]
\[=27\ast V\]
where V is Volume of sphere with radius 3
Since V = (4/3) pi*3^3 =36pi
Therefore
\[\iint_{\gamma}{\vec{F}.\hat{n}dS=27\ast36\ \pi}=972\pi\]