= Spatially averaged radiant exposure
{title2=$\overline H_A=Q_A/A$}
\b[Spatially averaged radiant exposure] is the received <radiant energy> divided by the specified receiving area. For a patch $A$ and local <radiant exposure> $H(\mathbf x)$,
$$
Q_A=\int_A H(\mathbf x)\,dA,\qquad
\overline H_A=\frac{Q_A}{A}.
$$
Both local and averaged values have units $\mathrm{J\,m^{-2}}$, but the average need not equal the local maximum. The <radiant exposure> definition uses incident energy and the receiving surface's area.<International Lighting Vocabulary by CIE>[Entry17-21-071, definition, symbol and unit]{ref}
For an original unequal-area example, let one $3\,\mathrm{mm^2}$ region receive a uniform $2\,\mathrm{J\,m^{-2}}$, and a neighbouring $1\,\mathrm{mm^2}$ region receive $10\,\mathrm{J\,m^{-2}}$. They receive $6$ and $10\,\mathrm{\mu J}$ respectively. The average over the combined patch is therefore
$$
\overline H_A=\frac{16\,\mathrm{\mu J}}{4\,\mathrm{mm^2}}=4\,\mathrm{J\,m^{-2}}.
$$
The unweighted average of the two listed exposure values would be $6\,\mathrm{J\,m^{-2}}$, which is wrong for these unequal areas. The <area weighting of radiant exposure> gives the larger region three times the influence. The actual peak remains $10\,\mathrm{J\,m^{-2}}$ in the stipulated piecewise-uniform model.
A receiver integrating all incident energy over the complete patch can determine $Q_A$; division by its known collecting area then gives the average. That measurement alone cannot identify the peak or reconstruct the spatial distribution. Subdivision with a characterized spatial response supplies additional information. For a finite aperture intercepting a <Gaussian beam>, use energy actually entering the aperture, as calculated in <radiant exposure of a Gaussian beam>, rather than the total energy of the untruncated beam.
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