Comment on So are they like extinct or something?
tetris11@feddit.uk 1 day agoit’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.
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 or higher curves (e.g. a pond ripple), 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, and creating a forgotten space just above the axis plane?
gandalf_der_13te@feddit.org 21 hours ago
well, almost. it’s a bit different than that.
what you mean is this:
Image
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:
Image
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.
tetris11@feddit.uk 20 hours ago
Ohh! Calculating area, not volume under the curve – I see, thank you