(It's mostly a calculus question and has little to do with optics.)
I'm reading a book on Computer Graphics, Realistic Image Synthesis Using Photon Mapping by Henrik Wann Jensen, and I can't completely understand how to derive the radiance equation given in Chapter 2.
According to the book, the spectral radiant energy $Q_{\lambda}$, in $n_{\lambda}$ photons with wavelength $\lambda$ is
$Q_{\lambda}=n_{\lambda}\frac{h \; c}{\lambda}\;,$
where h is Planck's constant. Hence
$Q=\int_{0}^{\infty}Q_{\lambda}d\lambda\;.$
Radiant flux $\Phi$ is the time rate of flow of radiant energy:
$\Phi=\frac{dQ}{dt}\;.$
And radiant flux area density is
$E(x)=\frac{d\Phi}{dA}\;.$
The radiant intensity $I$ is the radiant flux per unit solid angle $d\omega$:
$I(\omega)=\frac{d\Phi}{d\omega}\;.$
So the radiance $L$ is the radiant flux per unit solid angle per unit projected area:
$L(x,\omega)=\frac{d^{2}\Phi}{\cos\theta\;dA\;d\omega}=\int_{0}^{\infty}\frac{d^{4}n_{\lambda}}{\cos\theta\;d\omega\;dA\;dt\;d\lambda}\;\frac{h\;c}{\lambda}d\lambda$
My question is: Could anyone explain how to derive the last integral, and why it is not
$\int_{0}^{\infty}\frac{d^{3}n_{\lambda}}{\cos\theta\;d\omega\;dA\;dt}\;\frac{h\;c}{\lambda}d\lambda\;?$