Gauss's Divergence Theorem
This theorem gives relation between Surface Integral and Volume Integral.
The surface integral of the normal component of a vector function \(\vec{F}\) taken around a closed surface \(S\) is equal to the integral of the divergence of \(\vec{F} \) taken over volume \(V\) enclosed by the surface \(S\).
\[ \iint_S \vec{F}.\hat{N}ds = \iiint_V \vec{F}dV \]
Sample Problem
Usinf Gauss's Divergence Theorem, evaluate \(\hspace{5pt} \iint_S \vec{F}.\hat{N}ds \) if equation of the sphere \(S\) is \(x^2 + y^2 + z^2 = 16 \) and \(\vec{F} = 3x\hat{i} + 4y\hat{j} + 5z\hat{k} \).
Solution
By Gauss's Divergence Theorem, \[ \iint_S \vec{F}.\hat{N}ds = \iiint_V \vec{F}dV \] So calculating divergence, \[ \nabla .\vec{F} = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right).\left(3x\hat{i} + 4y\hat{j} + 5z\hat{k} \right) \] \[ = 3 + 4 + 5 = 14 \] Using the value of \(\nabla \vec{F}\) \[ \iint_S \vec{F}.\hat{N}ds = \iint_V \int14.dv \] \[ = 14V \] Since \(V\) is the volume of sphere, \[ V = \frac{4}{3}\pi r^3 \] \[ \text{So } 14V = 14 \times \frac{4}{3} \pi (4)^3 = \frac{3584\pi}{3} \] where 4 is the radius of the given sphere.
