I am trying to understand the divergence theorem, so I found a YouTube video by an MIT instructor online. There is something I don't understand, and I know it is something that should be easy and I should know it, and so I am wringing my hands trying to figure it out.
I am referring to 16:30-18:15 of this video:
Lec 28 | MIT 18.02 Multivariable Calculus, Fall 2007 - YouTube
The dude claims that the double integral on the surface of the circle of (x^2+y^2) dx dy is equal to the integral with limits 0 to 2pi and the integral with limits 0 to 1 of (r^2)*r dr dtheta, and this is equal to pi/2.
Somehow, this has to be right, but here is the part I don't get: I know that x^2+y^2=r^2, but I don't know why he multiplies by another r. In other words, why not "(r^2) dr dtheta" instead of "(r^2)*r dr dtheta"?
Usually at this stage of a course they insert the r just because it has to be there, since you get the wrong answer otherwise. The most concrete way to see this is to try to get the area of a circle using polar coordinates; if you just integrate 1 with theta from 0 to 2pi and r from 0 to R, you'll get 2piR, which is obviously not the area of a circle. If you put the r in, you get 2pi (R^2/2) = piR^2.
The actual reason this works is because when you make the coordinate transformation from cartesian to polar coordinates, you change the area of a small region of the space by a certain amount. That is, the area of a box of the form [a,a+da] x [b,b+db] in cartesian coordinates and a "box" of the from [r,r+dr] x [t,t+dt] are not the same.
This is easiest to see when the transformation is linear, such as mapping a coordinate system whose basis is [1,0], [0,1] into one whose basis is, say, [1,2], [3,1]. A 1x1 box in the first coordinate system is just a 1x1 square, which has area 1. A "1x1 box" in the second coordinate system is a parallelogram in the standard coordinate system. This can be seen by drawing the vertices [0,0], [1,2], [3,1], and [1+3,2+1]=[4,3] and connecting them to form a parallelogram. It is clear without even having to do any arithmetic that the area of this region is different from the area of the first region, despite the fact that the second region can be described as a 1x1 box in a coordinate system.
It is a theorem of linear algebra (no calculus required), that when you make a coordinate transformation from the standard basis [1,0,0...,0], [0,1,0,...], ..., [0,0,...,1] to a new basis v1,v2,...,vn, the "volume" of the 1x1x1x...x1 "box" in the new coordinate system is the determinant of the matrix whose rows (or columns) are v1,v2,...,vn. You may recall that the determinant that you get when you put two vectors in 2 dimensions into a matrix is the area of a paralleogram, while the same thing in 3 dimensions results in the volume of a parallelopiped. This generalizes that result, essentially.
When your region is not defined by a linear transformation but is instead defined by a curvy sort of transformation, like the one from Cartesian to polar coordinates, the idea is actually fairly similar: get the best linear approximation of the transformation at each point, and then multiply by the determinant of that matrix at each point, and add up the result with an integral in the usual way. This matrix corresponding to the best linear approximation is the Jacobian, and for the polar coordinate transformation matrix its determinant at (r,t) is r.
Intuitively, the area of the "box" [r,r+dr] x [t,t+dt] is r dr dt. Or, since theta is constant here, the area of an infinitely thin annulus extending from r to r+dr is 2pi r dr. This should begin to click now if it hasn't already: as you add up the circumferences of circles over varying radii beginning at 0, you fill out the area of a circle.
Incidentally, I really really wish someone would create a hybrid, perhaps year-long linear algebra+Calculus 3 course. So many things like this could be clearer this way.
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but it still didn't turn out right
