I found this mathematical physics question in an older exam and I was wondering how I could solve something like this because I have never seen such a question before:
$$\frac{d}{dt}\int_{a(t)}^{b(t)} \rho (0.5v^2 + χ) dx = \int_{a(t)}^{b(t)} \rho v b dx + \int_{a(t)}^{b(t)} (v\sigma)_xdx$$
Where $χ(x, t)$ is the internal energy density per unit mass.
In words, the above equation states that the rate of change of the total energy equals the sum of the rate of the work done by the external body forces $b$ and the rate of work done by the stress $σ$.
Now I have to show that the above expression results into the following equation:
$$\rho \frac{Dχ}{Dt} = \sigma \frac{\partial v}{\partial x}$$
Any explanation/hint/tip would be grateful. Thanks in advance.
We have that:
$\frac{d}{dt}\int_{a(t)}^{b(t)} \rho (0.5v^2 + χ) dx = \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} \rho (0.5v^2 + χ) dx$ and from here, I am stuck at what doing next.