this post was submitted on 08 Sep 2026
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oh they don't appear again, but its generalization, the parallelepiped, does:
it's used to calculate volumes of curved objects. basically you chop down the object into a lot of small parallelepipeds (mentally), and then calculate the volume of each of them small things and sum over them. Done.
to calculate the volume of a parallelepipede, there's a surprisingly simple mathematical formula. If you have the vectors for the three sides a, b, c, then the volume V = (a × b) · c, where × is the cross product and · is the dot product. it's very simple and an effective way to calculate volumes of curved / deformed objects.
Links:
pardon my ignorance but what the fuck is a "cross product" or a "dot product"? I assumed at first this was multiplication, but then I saw the dot and realized this isn't anything I've ever been taught
https://en.wikipedia.org/wiki/Cross_product
https://en.wikipedia.org/wiki/Dot_product
sorry i am too tired to explain in full detail rn
that's fine, and thanks for the links! I just don't quite agree that this is suprisingly simple mathematics :P
yeah it's "surprisingly simple" in the sense that there's a well-defined algorithm to do it. a computer can do it easily, with very little time/effort. anyways, you don't need to think about every problem specifically. "now i got this parallelepiped, how do i calculate the volume?" you can just use the same formula every time.
Sometimes I think about parallelepipeds then when I reassociate I'm smiling and my fiance is visibly wondering what I'm thinking about. "Don't worry hun, parallelepipeds again"
For anyone not quite getting this (like me), to calculate the area under a curve in 2D we were taught to cut it into tiny thin rectangles and sum those up; intergration when those rectangles have a width tending to 0.
For 3D (e.g. a pond ripple) or higher curves, a simple rectangle wont cut it as the length of the rectangle might only get the top of a wave, but not capture the crest of it tangential to it. So you create slopey rectangles to approximate that space, and bring the limit to zero to get the area (I think).
My only confusion now is, if I'm deforming a rectangle from one side to approximate the curve just above it, am I not also deforming the bottom of that rectangle in the same way (for the parralel strcture to hold true), and creating a forgotten space just above the axis plane?
well, almost. it's a bit different than that.
what you mean is this:
you approximate an integral with a lot of small thin rectangles, but if the curve's not entirely rectangular, there's gonna be some error, which becomes smaller as the rectangles become smaller. this can be ignored if the rectangles are thin enough. and it's not what i meant.
what i meant is something like this:
you take a piece of elastic fabric, and paint some squares on it. now, if you stretch the fabric, the squares change shape, they become approximately parallelepipeds. this is a good approximation. now, if we want to calculate the total area of that fabric (after stretching), we can calculate the area of each of the small parallelepipeds (which is easy to do with the formula in above comment) and then sum them.
Ohh! Calculating area, not volume under the curve -- I see, thank you