I'm pretty sure those integrals won't give the correct result.
I've been trying to wrap my head around this for a while. Visualizing this has been a challenge.
[itex]\displaystyle f(x,\,y,\,z)=\min(x,\,y,\,z)=\left\{\matrix{<br />
x\,,\ \ \text{ if }\ x\le y\ \ \text{ and }\ \ x\le z\\<br />
y\,,\ \ \text{ if }\ y\le x\ \ \text{ and }\ \ y\le z\\<br />
z\,,\ \ \text{ if }\ z\le x\ \ \text{ and }\ \ z\le y}\right.[/itex]
My suggestion is to write f(x, y, z) as the sum of three functions, f
x(x, y, z) + f
y(x, y, z) + f
z(x, y, z), where
fx(x, y, z) = x when x is the minimum, otherwise it equals zero.
fy(x, y, z) = y when y is the minimum, otherwise it equals zero.
fz(x, y, z) = z when z is the minimum, otherwise it equals zero.
Then integrate all three functions over the entire cube. But since each is zero over 2/3 of the cube, simply integrate each over the region for which it's not zero. Considering symmetry, each of those integrals will give the same result.
For example, integrate y first for x going from y to 1, then for z going from y to 1, then for y going from 0 to 1.
[STRIKE]Added in
Edit:
I struck part of the line above after looking at the figure in
jackmell's post. That line should read:
For example, integrate y first for x going from y to 1, then for z going from 0 to y, then for y going from 0 to 1.
[/STRIKE]
Added in
2nd Edit:
After further consideration, I withdraw the 1st Edit. and restore the post as it was first written. (This does not imply that
jackmell's post was wrong -- only that I misinterpreted his figure.)
