# Poynting vector flux in a tube (geometrical optics)

Let's consider the following tube (defined as the solid described by two arbitrary surfaces dA1 and dA2, and by all the rays which connect their boundaries):

I read in a textbook that, because of Poynting theorem, the flux of the Poynting vector through all the tube is necessarily 0 (and then this fact is used to prove that, since the flux through the lateral surface is 0, the flux through dA1 is the same of that through dA2). But I do not understand the first statement.