= Radiant exitance
{title2=$M=M_e=d\Phi_{\rm emitted}/dA$}
{wiki}
<Radiant exitance> is total outward emitted <radiant flux> per actual emitting <area>, measured in $\mathrm{W\,m^{-2}}$. Integrate <radiance> over the outward <hemisphere>, keeping the <projected area> factor:
$$
M=\int_{\Omega_{\rm outward}}L(\theta,\varphi)\cos\theta\,d\Omega.
$$
If the <radiance> is direction-independent, the <spherical coordinates> element is $d\Omega=\sin\theta\,d\theta\,d\varphi$, so
$$
M=L\int_0^{2\pi}\!d\varphi\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\pi L.
$$
The weighted <solid angle> is $\pi\,\mathrm{sr}$, although the whole <hemisphere> covers $2\pi\,\mathrm{sr}$. Directions near grazing contribute less because of the <projected area>. This geometric factor is the bridge between <Lambert's cosine law> and the total output of a diffuse patch. https://labs.phys.utk.edu/mbreinig/phys421/modules/m5/Radiometry.html[Breinig, Radiometry: hemispherical integration of radiance]{ref}.
For the uniform patch with $L=10\,\mathrm{W\,m^{-2}\,sr^{-1}}$, <radiant exitance> is $M=10\pi\,\mathrm{W\,m^{-2}}$. Its <area> $0.020\,\mathrm{m^2}$ gives total <radiant flux> $\Phi=MA=0.20\pi\,\mathrm W\approx0.628319\,\mathrm W$ over the outward <hemisphere>. This total differs from its <radiant intensity> $I(0)=0.20\,\mathrm{W\,sr^{-1}}$, which describes only one direction.
For an ideal <blackbody>, integrate <Planck's law> over <frequency> as well: $M=M_e=j^*=\sigma T^4$ by the <Stefan–Boltzmann law>. An ideal $6000\,\mathrm K$ emitting patch of <area> $1\,\mathrm{mm^2}$ emits about $73.4881\,\mathrm W$ outward. <Irradiance> counts incoming radiation separately; for purely emitted output the net outward exchange per <area> is $M-E$. Reflected output must also be included if the surface reflects incident radiation.
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