I have I believe a simple question which relates to two-dimensional integral. I have been reading enough (and watch video tutorials) on the technique to hopefully understand it well. Every tutorial takes the assumption that we measure somehow the "volume" under a surface when we deal for instance with two-dimensional integrals:
$z = \int_x \int_y f(x,y)dx dy$
Let me know if I get that wrong, but I think it is a generally good answer. The problem I have in this. I am trying to understand two-dimensional integrals within a different context. More precisely I am reading a book on computer graphics, in which they say that pixels of an image represent radiance, and that this radiance needs to be computed with an integral over the pixel's area. Something like this:
$L_{pixel} = \int_{pixel area} L(x)dx$
where L stands for radiance. I understand this as well more or less, but then if a two dimensional integral applied to a surface gives a volume, what is the unit of a two-dimensional integral applied to radiance over the area of pixel? I would expect radiance as well, but since the integral of a surface give cubic meters, shouldn't it be something else than radiance? Cubic radiance? What is the meaning of the value returned?
I am sure I am just being confused but would really appreciate if someone could clarify this for me.
Thank you.
In general, the units of $\iint f(x,y)\,dx\,dy$ are the units of $f$ times the units of $x$ times the units of $y$. If $x,y$ have units of length, then in order of the integral to make sense as Quantity of X, the units of $f$ must be "X per square meter", or simply density of $X$.
Here's another version: $\iiint f(x,y,t)\,dx\,dy\,dt$ where $x,y$ are as above and $t$ is time. Here $f$ could be flux, i.e., flow rate per unit area. This is measured in, say, kilograms per square meter per second. Then the integral is measured in kilograms: it's the total amount that passed through the given area in given time interval.