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Proof of Identity for Volume Integral of Powers of r

Theorem
\[\int \int \int_V nr^{n-2} \mathbf{r} dV = \int \int_S f^n \mathbf{n} dS\]
  for a region  
\[V\]
  bounded by a surface  
\[S\]
.
Proof
\[\begin{equation} \begin{aligned} \mathbf{\nabla} r^n &= (\frac{\partial}{\partial x} \mathbf{i} + \frac{\partial}{\partial y} \mathbf{j} + \frac{\partial}{\partial z} \mathbf{k}) (x^2 +y^2 +z^2)^{n/2} \\ &= nx(x^2 + y^2 +z^2)^{(n-2)/2} \mathbf{i} + ny(x^2 + y^2 +z^20^{(n-2)/2} \mathbf{j} + nz(x^2 + y^2 +z^20^{(n-2)/2} \mathbf{k} \\ &= n(x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) r^{(n-2)/2} \\ &= nr^{n-2} \mathbf{r} \end{aligned} \end{equation}\]

Now use the  
\[\int \int \int_V \mathbf{\nabla} \phi dV = \int \int_S \phi \mathbf{n} dS\]
  with  
\[\phi = r^n\]
  to give
\[\int \int \int_V nr{n-2} \mathbf{r} dV = \int \int_S f^n \mathbf{n} dS\]
  for a region  
\[V\]
  bounded by a surface  
\[S\]