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I'm struggling to find a mathematical expression for the total fluence of a traveling electromagnetic wave (through vacuum).

Though its not really a physics problem rather than a problem with basic math, I guess...

I'm pretty sure that I'm just being extraordinarily stupid...

As an example, consider the following sketch of a unit volume with a unit surface area $A$. The wave travels to the right.

sketch 1, axes are labeled in 2nd sketch

Now suppose, that we know the energy density $u(t)$ at every instant in time. Then the energy flux $\Phi(t)$ being the energy $E(t)$ crossing the area $A$ per unit time $\Delta t$ is simply $$\Phi(t)=\frac{E(t)}{A\Delta t}=u(t) c_0$$ where $c_0$ is the vacuum speed of light.

The total energy, which has crossed the same area within some time-interval $t_1-t_0$ is then the time-integral of the above expression

$$F=\int_{t_0}^{t_1} u(t)c_0 \,\mathrm dt ,$$

having units of $\mathrm{J/m²}$

Now consider the energy density to be a function of time and $y$-coordinate, so that every volume-element $\,\mathrm dA\cdot c_0\,\mathrm dt$ (2nd sketch) contains a different amount of energy

2nd sketch, inhomogeneous distribution of energy-density

The total energy within each "slice" is $$E(t,y)=u(t,y)\cdot \,\mathrm dA\cdot c_0 \,\mathrm dt$$

This is as far as I can get... How can I find an equivalent expression for the total fluence $F$ as above? (Not just $F(y)$ but the total $F$, i.e. integrated along $y$ as well)

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As an example, consider the following sketch of a unit volume with a unit surface area $A$.

If $A$ is a unit area, the volume isn't equal to $1$, but to $c\,\mathrm d t$.

Then the energy flux $\Phi(t)$ being the energy $E(t)$ crossing the area $A$

This should read

Then the energy flux $\Phi(t)$ being the energy $E(t)$ crossing the unit area

Your question is:

How can I find an equivalent expression for the total fluence $F$ as above?

The total energy per unit area is still

$$F(y)=\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt$$

Let $z_0$ be the thickness of the cuboid and write $\mathrm dA$ as

$$\mathrm dA = z_0 \, \mathrm dy$$

The total energy per area $\mathrm dA$ is therefore

$$F_{\mathrm dA} = F(y)\,\mathrm dA = \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right) (z_0 \, \mathrm dy)$$

Now you can get the net energy passing through the entire unit right side by integrating from $0$ to $y_0$ where $y_0$ the height of the cuboid.

$$F = \int\limits_{0}^{y_0} F(y)\,\mathrm dA = \int\limits_{0}^{y_0} \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right)\,z_0\,\mathrm dy$$

Since the right side is a unit area,

$$z_0y_0 = 1\Rightarrow y_0 = \frac 1 {z_0}$$

Therefore

$$F = \int\limits_{0}^{\frac 1 {z_0}} \left(\int\limits_{t_0}^{t_1} u(t,y)c_0 \,\mathrm dt\right)\,z_0\,\mathrm dy$$

$F$ is the net energy passing through the right side (of the unit area) of the cuboid.

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