Equation of Continuity of charge for point charges

[transience]

Homework Statement

I am looking to demonstrate that the expressions for the charge and current density of point charges satisfy the equation of continuity of charge. Intuitively it makes sense to me but I run into trouble with the delta function when I try to prove it mathematically.

Homework Equations

\rho(\vec{x},t)=\sum_i q_i \delta (\vec{x}-\vec{x}_i(t))

\vec{j}(\vec{x},t)=\sum_i q_i v_i(t) \delta (\vec{x}-\vec{x}_i(t))

\frac{\partial}{\partial t}\rho(\vec{x},t)=- \nabla \cdot \vec{j}(\vec{x},t)

\delta^{3}(\vec{x}-\vec{x}')=\frac{1}{r^2\sin\theta} \delta(r-r') \delta(\theta-\theta')\delta(\phi-\phi')

The Attempt at a Solution

I start by constructing a sphere of radius a centered at \vec{x}=0

Using the integral form of the equation of continuity of charge in spherical coordinates I get
\frac{\partial}{\partial t}\int\int\int_V \rho(\vec{x},t) dV=-\int\int_S \vec{j}(\vec{x},t)dS

Then
LHS=\frac{\partial}{\partial t}\int\int\int_V \rho(\vec{x},t) dV
LHS=\frac{\partial}{\partial t}\int_0^{2 \pi} \int_0^\pi \int_0^a \sum_i q_i \frac{1}{r^2\sin\theta} \delta(r-r_i(t)) \delta(\theta-\theta_i(t))\delta(\phi-\phi_i(t)) r^2\sin{\theta} dr d \theta d \phi

Working the maths through gives me
LHS=\sum_i q_i \frac{\partial}{\partial t}\int_0^a \delta(r-r_i(t)) dr

Then I do basically the same thing on the right hand side
RHS=-\int_0^{2 \pi}\int_0^{\pi} \sum_i q_i v_{ir}(t) \frac{1}{r^2\sin\theta} \delta(r-r_i(t)) \delta(\theta-\theta_i(t))\delta(\phi-\phi_i(t)) a^2\sin{\theta} dr d \theta d \phi

Which works out to be

RHS= \sum_i q_i v_{ir}(t) \frac{a^2}{r^2} \delta(r-r_i(t))

I can't figure out where to go from here, any help would be greatly appreciated, thanks.

 

[sgd37]

using a change of variable

\delta (r-r_i(t)) = \delta (r'(t))

your LHS becomes

\sum_i q_i \frac{\partial r'(t)}{\partial t}\int_0^a \frac{\partial}{\partial r'} \delta(r'(t)) dr' = \sum_i q_i v_i (t) \delta (r - r_i (t) )

as to the factors of a/r I'm not sure