From my course in mathematical physics, we define the function: $$e(t)=\int_{B_{c(t_0-t)}(x_0)}dx\left[\frac{1}{2}\left(\frac{\partial u}{\partial t}\right)^2+\frac{c^2}{2}|\nabla u|^2 +\frac{\gamma}{2}u^2\right]$$ With $t\in[0,t_0]$ and $B_{c(t_0-t)}(x_0)$ being the ball centered in $x_0$ with radius $c(t_0-t)$.
We wish to calculate $\frac{d}{dt}e(t)$, the notes go as follows: $$\frac{d}{dt}e(t)=\frac{d}{dt}\int_{0}^{c(t_0-t)}dr\int_{\partial B_{r}(x_0)}d\sigma\left[\frac{1}{2}\left(\frac{\partial u}{\partial t}\right)^2+\frac{c^2}{2}|\nabla u|^2 +\frac{\gamma}{2}u^2\right]$$ $$=-\frac{c}{2}\int_{\partial B_{c(t_0-t)}(x_0)}d\sigma\left[\left(\frac{\partial u}{\partial t}\right)^2+c^2|\nabla u|^2 +\gamma u^2\right] + \int_{B_{c(t_0-t)}(x_0)}dx\left[\frac{\partial u}{\partial t}\left(\frac{\partial^2 u}{\partial t^2}+\gamma u\right)+c^2\nabla u\cdot\nabla\left(\frac{\partial u}{\partial t}\right)\right]$$ Etc, etc... I find myself slightly unsure about the procedure followed to differentiate, I understand the first term $-\frac{c}{2}\int_{\partial B_{c(t_0-t)}(x_0)}d\sigma\left[\left(\frac{\partial u}{\partial t}\right)^2+c^2|\nabla u|^2 +\gamma u^2\right]$ as the integrand of the $dr$ integral evaluated in its upper extremum $c(t_0 - t)$, where does the second term come from? the: $$\int_{B_{c(t_0-t)}(x_0)}dx\left[\frac{\partial u}{\partial t}\left(\frac{\partial^2 u}{\partial t^2}+\gamma u\right)+c^2\nabla u\cdot\nabla\left(\frac{\partial u}{\partial t}\right)\right]$$ I can see that the integrand here is just the derivative with respect to time of the expression in the original integral, I however am not sure as to where does this term come from, and why we integrate over the whole ball.
Given an integral of the form $$\Phi(t)=\int_{\Omega(t)}F(t,\boldsymbol x)\mathrm d^3\boldsymbol x$$ In order to calculate $\Phi'(t)$ you need to use the Reynolds transport theorem, $$\Phi'(t)=\int_{\Omega(t)}(\partial_t F)(t,\boldsymbol x)\mathrm d^3\boldsymbol x+\int_{\partial\Omega(t)}\big((\boldsymbol v\cdot\boldsymbol n)F\big)(t,\boldsymbol x)\mathrm d^2\boldsymbol x$$ Where $\Omega(t)$ is a "continuous" set-valued function (this is difficult but not impossible to define rigorously) and $\boldsymbol v(t,\boldsymbol x)$ is the "velocity of the boundary", which again in general is hard to define rigorously, but when $$\Omega(t)=\mathbb B(\boldsymbol x_0,c(t_0-t))$$ It's pretty clear that $$\boldsymbol v(t,\boldsymbol x)=-c~\frac{\boldsymbol x}{|\boldsymbol x|}$$ This is where the mysterious surface integral comes from. The volume integral just comes from the Leibniz rule.