Find the Divergence of the Vector Field $A(x,y,z)=\frac{1}{(x^2+y^2+z^2)^\frac{3}{2}}(x,y,z)$

Question

We need to find the Divergence of the Vector Field $A(x,y,z)=\frac{1}{(x^2+y^2+z^2)^\frac{3}{2}}(x,y,z)$ and show that it is $0$ apart from the origin. The part of this problem that confused me was the way the vector field was written.

Answer 1

First, we start by rewriting the vector field as $A(x,y,z)=(\frac{x}{(x^2+y^2+z^2)^\frac{3}{2}},\frac{y}{(x^2+y^2+z^2)^\frac{3}{2}},\frac{z}{(x^2+y^2+z^2)^\frac{3}{2}})$

Now we have to find the partial derivatives $\frac{\partial A}{\partial x}$, $\frac{\partial A}{\partial y}$, and $\frac{\partial A}{\partial z}$.

By observing the differentiation rules we find the following:

• $\frac{\partial A}{\partial x}=\frac{(x^2+y^2+z^2)^\frac{3}{2}-3x^2(x^2+y^2+z^2)^\frac{1}{2}}{(x^2+y^2+z^2)^3}$

• $\frac{\partial A}{\partial y}=\frac{(x^2+y^2+z^2)^\frac{3}{2}-3y^2(x^2+y^2+z^2)^\frac{1}{2}}{(x^2+y^2+z^2)^3}$

• $\frac{\partial A}{\partial z}=\frac{(x^2+y^2+z^2)^\frac{3}{2}-3z^2(x^2+y^2+z^2)^\frac{1}{2}}{(x^2+y^2+z^2)^3}$

Lastly we add our partial derivatives in order to obtain the divergence of the Vector field:

$divA=\frac{\partial A}{\partial x}+\frac{\partial A}{\partial y}+\frac{\partial A}{\partial z}=\frac{3(x^2+y^2+z^2)^\frac{3}{2}-3(x^2+y^2+z^2)^\frac{1}{2}(x^2+y^2+z^2)}{(x^2+y^2+z^2)^3}=\frac{3(x^2+y^2+z^2)^\frac{3}{2}-3(x^2+y^2+z^2)^\frac{3}{2}}{(x^2+y^2+z^2)^3}=0$