Hyperconcise Math F5: Integrals over space

Disclaimer: this topic is far less general than the rest of this series. I’m including it here anyway because the general case is complicated, but still knowable, and an annoyingly drawn out yet incomplete overview of these specifics is a huge part of a standard calculus curriculum.

If f(x, y, z) = 1, the integral equals the total volume of the region.

To change the variables (x, y, z → u, v, w) in a volume integral, rewrite the equations for the boundary in terms of the new variables (and arrange them so that the integral is positive if f(x, y, z,) = 1), then replace f(x, y, z) dz dy dx with f(x(u, v, w), y(u, v, w), z(u, v, w))⦁|J| dw dv du. J is a number that accounts for any curvature introduced by the new coordinates (so the volume of the space does not change):

The det() function comes from linear algebra, and the change of coordinates must be one-to-one (so J ≠ 0 everywhere).

For example,

All graphs here were made using Geogebra (slightly modified), under CC BY-SA 3.0

These integrals represent how much f(x, y, z) is contained in (or how much g(x, y, z) is aligned with) the curve or surface defined by r (the boundary is defined by u = u₂, u = u₁, v = v₂, v = v₁):

If f(x, y, z) = 1, the integral equals the total length of the curve or area of the surface. For the curve or surface to be well defined, r⁽ᵘ⁾ and r⁽ᵛ⁾ should not equal 0 anywhere.

For example,

Like the fundamental theorem of calculus, integrals of vector functions over manifolds (generalized surfaces) can be found using just information on the boundary with the generalized Stokes’ theorem. Since a complete understanding of this theorem requires familiarity with differential forms, here are some common special cases instead:

X, Y, and Z are “dummy” variables used to distinguish the coordinates you are integrating over from the coordinates in the value of the integral.