Since the hyper-volume of a hypercube of side L is L^4 and the surface volume
is 8L^3, what about a hyper rectangular box? I can see the surface volume would
be 8(L*W*H) but what would the hyper volume be?
Yup, that's why it's a hyper box, it is 4 dimensional. But with the hypercube
(tesseract) it's easier to see because each of the hypercube's edges is the
same length, so there's no confusion that hyper volume is L^4. But with the
hyper rectangular box each of the edges are a different length so which one
do you use for the hyper volume?
From what I understand, each 'surface' of a hypercube is a 3D cube. So if it's
not a hypercube but a hyper rectangular box then each 'surface' of it would be a
3D rectangular box. And if L, W and H are each different... that's where my
confusion comes in about the hyper volume.
Romsek answered that confusion. For surface volume, it would be $2(L\cdot W \cdot H + L\cdot W\cdot H' + L\cdot H\cdot H' + W\cdot H\cdot H')$. When $L = W = H = H'$, you get the special case: $2(L^3+L^3+L^3+L^3) = 8L^3$. For hyper volume, you have $L\cdot W\cdot H\cdot H' = L^4$.
Hmm, ok I think. Let's try an example.
If I have a 3D box that is 3 by 5 by 9 and I translate it up to the 4th
dimension, the L, W and H will still be 3, 5 and 9, yes? So then what
will the H' be? Or do I not translate up but assign the edges of the
4D box, say 3 by 5 by 9 by 11?
"Picturing" higher dimensional objects is typically easiest with simple objects, like simplices:
https://en.wikipedia.org/wiki/Simplex
Short answer: yes.
Longer answer: Typically, negative measures require a generalization of volume that result from Lesbesgue integration over a signed measure, which itself is a generalization of the traditional nonnegative measure. I will not go into much more details on it, as Measure Theory is a rather complex subject already.