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Reynold's Transport Theorem states that

\[\frac{d}{dt} \int \int \int_{V_t} f dV =\int \int int_{V_t} \frac{\partial f}{\partial t} dV + \int \int_{S_t} f \mathbf{v} \cdot d \mathbf{S}\]

where  
\[f\]
  is a function of  
\[\mathbf{r}, \:t\]
  defined on a volume  
\[V\]
  consisting of a sphere expanding at a constant speed  
\[v\]
  so that  
\[r=vt\]

The Continuity Equation states  
\[\frac{\partial \rho}{\partial t} + \mathbf{\nabla} (\rho \mathbf{v}) =0\]

Set  
\[f = \rho\]
  in Reynold's Theorem gives
\[\frac{d}{dt} \int \int \int_{V_t} \rho dV =\int \int int_{V_t} \frac{\partial \rho}{\partial t} dV + \int \int_{S_t} \rho \mathbf{v} \cdot d \mathbf{S}\]

Now apply the Divergence Theorem  
\[\int \int_S \mathbf{F} \cdot d \mathbf{S} = \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV\]

with  
\[\mathbf{F} = \rho \mathbf{v}\]
  to get
\[\frac{d}{dt} \int \int \int_{V_t} \rho dV =\int \int \int_{V_t} \frac{\partial \rho}{\partial t} dV +\int \int \int_{V_t} \mathbf{\nabla} \cdot (\rho \mathbf{v}) dV\]

Hence  
\[\frac{d}{dt} \int \int \int_{V_t} \rho dV =\int \int \int_{V_t}( \frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot (\rho \mathbf{v})) dV=0\]

Then the left hand side is zero and  
\[ \int \int \int_{V_t} \rho dV =constant\]