Energy in Electromagnetic Waves Looking at diagrams of Electromagnetic Waves, it would appear to me that at certain times the waves have zero amplitude, and consequently zero energy. Indeed, substituting in the sinusoidal terms into the Poynting Vector equation, It would seem that at certain times the energy disappears. Why is this not the case?
 A: At certain positions in the waves, the EM field is zero and thus zero energy is stored at those positions. But at other positions, the EM field is at a maximum, and those points are local maxima of energy. That pattern of oscillation between zero energy and maximum energy moves in the direction of propagation of the wave but never changes - in particular, the maximum value of the EM field (the amplitude) stays constant, and there is no time at which the EM field is zero everywhere.
As for your conclusion from the definition of the Poynting vector that the energy disappears at certain times: it's not correct, but I couldn't tell you why without seeing how you did it. What I can do is show the calculation for an electromagnetic plane wave, defined by
$$\vec{E}(z,t) = E_0\hat{x}\sin(kz - \omega t)$$
The corresponding magnetic field is
$$\vec{B}(z,t) = \frac{1}{c}\hat{k}\times\vec{E}(z,t) = \frac{E_0}{c}\hat{y}\sin(kz - \omega t)$$
since I'm setting the direction of propagation as $\hat{k} = \hat{z}$. Check that this satisfies Maxwell's equations if you want. The energy density is
$$\begin{align}
u(z,t)
&= \frac{\epsilon_0}{2}E(z,t)^2 + \frac{1}{2\mu_0}B(z,t)^2 \\
&= \frac{\epsilon_0}{2}\biggl(E_0\hat{x}\sin(kz - \omega t)\biggr)^2 + \frac{1}{2\mu_0}\biggl(\frac{E_0}{c}\hat{y}\sin(kz - \omega t)\biggr)^2 \\
&= \epsilon_0 E_0^2\sin^2(kz - \omega t)
\end{align}$$
using $\frac{1}{c^2} = \epsilon_0\mu_0$. This energy density does vary from point to point, but at any fixed time, if you take the average over one cycle, a length $\frac{2\pi}{k}$, you get
$$k\int_0^{2\pi/k} u(z,t)\mathrm{d}z
= k\int_0^{2\pi/k} \epsilon_0 E_0^2\sin^2(kz - \omega t)\mathrm{d}z
= k\epsilon_0 E_0^2 \frac{\pi}{k}
= \pi\epsilon_0 E_0^2$$
which does not depend on time. So the average energy density is constant, it does not ever go to zero.
A: "Looking at diagrams of Electromagnetic Waves, it b would appear to me that at certain times the waves have zero amplitude, and consequently zero energy. Indeed, substituting in the sinusoidal terms into the Poynting Vector equation, It would seem that at certain times the energy disappears. Why is this not the case?" Anthony.
You are right Anthony, the electromagnetic wave is depicted by Maxwell as a wave that is changing its energy in time, as if the energy is going into a reservoir out of space and coming back.
"If the E and B both have sin components, the Poynting vector should have a sin^2, shouldn't it? This should be equal to zero at some point-shouldn't it? "
– 
Yes, Anthony, as the electric and the magnetic fields are the only place where the energy is stored in the wave, and these fields are in phase (E=cB) reaching its maxima at the same time, at the points you refer, the value of the energy, that is the sum of the energy contained in E and B will be equal to zero. 
The Pointing  Vector is defined as the velocity  of the energy flux per unit of area, 
S = 1/µ  ExB,  which implies again a variation of the energy. To face  the problem of the changing energy, another  concept  is  utilized, the Mean Pointing Vector (Sm = 1/2µO   ExB).  The Mean Pointing vector is defined as the mean of the velocity of the energy flux per unit of area,  and  is calculated when the electric and magnetic fields  reach their maxima.
“As for your conclusion from the definition of the Poynting vector, that the energy disappears at certain times: it's not correct, but I couldn't tell you why without seeing how you did it. What I can do is show calculation for an electromagnetic plane wave, defined by
E ⃗ (z,t)=E 0 x ^ sin(kz−ωt) 
The corresponding magnetic field is
B ⃗ (z,t)=1c k ^ ×E ⃗ (z,t)=E 0 c y ^ sin(kz−ωt) 
Since I'm setting the direction of propagation as k ^ =z ^. Check that this satisfies Maxwell's equations if you want. The energy density is
u(z,t) =ϵ 0 2 E(z,t) 2 +12μ 0  B(z,t) 2 =ϵ 0 2 (E 0 x ^ sin(kz−ωt)) 2 +12μ 0  (E 0 c y ^ sin(kz−ωt)) 2 =ϵ 0 E 2 0 sin 2 (kz−ωt) using 1c 2  =ϵ 0 μ 0. This energy density does vary from point to point, but at any fixed time, if you take the average over one cycle, a length 2πk, you get:
k∫ 2π/k 0 u(z,t)dz=k∫ 2π/k 0 ϵ 0 E 2 0 sin 2 (kz−ωt)dz=kϵ 0 E 2 0 πk =πϵ 0 E 2 0, which does not depend on time. So the average energy density is constant, it does not ever go to zero”. David Z.
David, what I am telling you is exactly what you said, that the energy of the electromagnetic wave at some point is equal to zero and it reaches to a maximum at other points. It is clear that if you take the whole cycle 2πk we will have the averaged energy.  And with it, the impression that the energy is conserved. But here comes the question: why there is a need to average an energy that is constant and must be the same at every moment? 
The logical conclusion is: when Maxwell’s equations are applied to the light wave we have a violation to the principle of conservation of energy. That means that at least, electromagnetism is incomplete, because it cannot describe the electromagnetic wave appropriately.
