Integral over a solid angle

I've been reading about energy conservation and radiosity from the perspective of computer graphics. The basic idea is simple enough: For all possible incoming light directions $\vec{l}$ and view directions $\vec{v}$ $\text{f}$ ($\vec{l}$, $\vec{v}$) should always be less than or equal to 1, where $\textbf{f}$ is the bidirectional reflectance distribution function (defines how light reflects off of a surface).

What I do not understand at all is how this is expressed mathematically. For example, Figure 3 on Page 12 of this presentation: http://blog.selfshadow.com/publications/s2012-shading-course/hoffman/s2012_pbs_physics_math_notes.pdf

$$\forall \vec{l},\int_\Omega f(\vec{l}, \vec{v})(\vec{n} \cdot \vec{v})d\omega_o \leq 1$$

First, I'm not clear on the notation. I think $\Omega$ indicates I should be taking the integral over a surface (in this case, a hemisphere) but I can't begin to actually solve it (or even input it into Mathematica). I have seen it written as a double integral using polar coordinates, which looks a little less mysterious though: http://www.rorydriscoll.com/2009/01/25/energy-conservation-in-games/. Perhaps they're equivalent?

But, more importantly, I'm lacking the intuition to see how any integral would be able to show what this one purports to show. I'm used to thinking of the integral as area or volume and I'm failing to see how that relates to anything in this case. Appreciate any insight.

 Paul Sinclair answered the second half of my question quite well but I'd still like to know how one might go about solving the given integral. I could give $\text{f}$ a definition but there's never going to be a $\omega$ anywhere in the integrand. Nor have I ever seen an integral that involves vectors. Is this some kind of math-pseudo code? Am I just expected to, for example, know that I should rewrite the problem in terms of polar coordinates, similar to what was done in the second link above? Replacing dot products with cosine terms? This isn't something I can just punch into mathematica, right?

Apologies for the scattered questions. I assure you I have spent time trying to answer them on my own, but I still feel like I'm somewhere far from the calculus 1-3 track I was somewhat familiar with. And so I'm not quite sure how to get from where I am to being able to understand everything I'd like to. Especially when notation/convention tends to vary depending on source. Even some search phrases would be extremely helpful.

Integration is more than just "area or volume". It is a sort of generalized sum. When you integrate a function, the result can be thought of as the sum of the function values over the region of integration.

I am going to have to base this on some guesses, as your link is far too fat for my connection. But yes, $\Omega$ is a surface, $\vec{n}$ represents the normal vector to the surface at the point $\vec{v}$ of integration and I am guessing that $\text{d}\omega_o$ is the differential of surface area (though that is a very odd notation for it). What the integral is saying is that if you sum up all the light that is reflected out in all directions from the point of reflection, that total amount of light is less than or equal to the amount of light shining onto the point. I.e, reflection does not increase the amount of light available.

• That is a much more general way to think of the integral, thank you. And given that, I think I can see how that would be the area under a curve. Evaluating the function over the entire region of integration just gives me the average "height" then? Sep 18, 2015 at 14:41
• Actually, we define the average of the function to be the ratio of the integral over the region to the measure (length, area, volume, whatever) of the region. Sep 18, 2015 at 22:19