1.2.6 Euler equations: from integral to differential form I

To transform the mass transport equation into differential form the first line of Eq. (15) is divided by $\left( t_1 - t_0 \right)$,

$$\int_{V} \frac{ \rho \left( \ensuremath{\mathchoice{\mbox{\... ...math$\scriptscriptstyle n$}}}} \; d\!A dt = 0 \enspace .$$ (16)

Taking the limes $t_1 \rightarrow t_0$ and assuming that the derivative $\frac{\partial \rho}{\partial t}$ exists for all $\ensuremath{\mathchoice{\mbox{\boldmath$\displaystyle x$... ...ox{\boldmath$\scriptstyle x$}} {\mbox{\boldmath$\scriptscriptstyle x$}}}$ we get

$$\int_{V} \frac{\partial \rho}{\partial t} dV + \oint... ...{\boldmath$\scriptscriptstyle n$}}}} \; d\!A = 0 \enspace .$$ (17)

Now, the Gauß theorem is applied (assuming that the divergence $\ensuremath{\mathchoice{\mbox{\boldmath$\nabla$}} {\mbox... ...ox{\boldmath$\scriptstyle v$}} {\mbox{\boldmath$\scriptscriptstyle v$}}}$ does exist) to transform the surface integral into a volume integral. We get

$$\int_{V} \frac{\partial \rho}{\partial t} dV + \int_... ...ox{\boldmath$\scriptscriptstyle v$}}} dV = 0 \enspace .$$ (18)