If the side of a cube comes from a uniform distribution between 1 and 5 inches, what's the expected volume of the cube ?The answer is to get the integral of x^3 at 5 minus the integral at one divided by 5-1. The integral is (x^4)/4 so we get 1/4(625/4-1/4) = 39.

But what if we are dealing with a cuboid and the sides are independent random variable. Well its E(length) * E(breadth) * E(width) or 3*3*3. But I wasn't sure if I was correct so I went to one of the newer tools in my mathematical toolbox - that's right I went triple integral.

It's the same as the integral above except we need to remember to divide by 64 because we are working against three axes.

Yay maths works!