The SIXTH of the seven beauties. The integral of the position vector dotted with the unit normal.

[@benhsu42]

Thank you! These have all been beautiful

[@MathTheBeautiful]

Thank you! I appreciate you!

[@harryharperable]

Yes exactly @APaleDot To rephrase but only using concepts introduced in the course: Since we know the integral doesn’t depend on the choice of origin, let’s choose a convenient one - an arbitrary point inside the closed curve. Now consider the infinitesimal triangle spanned by R(s) and R(s+ds). The “base” of the triangle has length ds because this is an arc length parametrisation. The “height” is by definition the projection of R in the direction of N. Since (T,N) form an orthonormal basis, R can be written as (R•N)N + (R•T)T, which shows that the height is R•N. Therefore the area enclosed within the curve is the sum of the areas of all the infinitesimal triangles, ie the integral of 1/2*R•N dL And so the beauty is shown* - Integral R•N dL = 2*A.

[@harryharperable]

*Technically I’ve only shown the case where the closed curve forms the boundary of a radially convex / “star domain” (a region where an origin exists such that all R vectors are contained within the region). It’s a bit hand wavy but I suspect we might be able to “cut” a more general region (?with some special property still) into a series of star regions and add up the integrals. The integral of R*NdL on the internal “cut” segments would then cancel out to zero because the orientation of N will be opposite in the corresponding segments of any pair of regions formed by a cut. The sum of the integrals on each sub-curve would then be (twice) the sum of the areas and would also be equivalent to the integral on just the overall boundary.

[@APaleDot]

@@harryharperable I didn't mention it in my comment because I didn't want to add burdensome complications, but my method works for any shape domain, and for any position of the origin. This happens because R•N computes the _signed_ area of the parallelogram between R and T. When R is aligned with N, the area is positive. When they're pointed in opposite directions the area is negative. The upshot is that sweeping R clockwise relative to the origin gives the opposite sign as sweeping anti-clockwise. And so if the boundary of the domain is not radially convex, there are portions where R must sweep in the opposite direction and this cancels out any area that gets swept outside the boundary.

[@APaleDot]

Near the beginning, you asked for a way to evaluate this integral using geometric reasoning and I think I have the answer for you: If you were to sweep R around the curve, it would obviously sweep out the area within the curve. We can approximate this using arbitrarily many triangles whose bases approximate the curve and whose sides are R(s) and R(s + ds). Taking the integral would mean letting the amount of triangles approach infinity, and then the bases of the triangles approach Tds. It's a well known fact in geometry that you can compute the area of the parallelogram swept out between two vectors by rotating one of them by 90° and taking their dot product. This is sometimes called the 2D cross-product, or more generally the exterior/wedge product. This is precisely the situation you're in where N is the rotated version of T and therefore R·N represents the parallelogram swept out between R and T. Given that a triangle is simply a parallelogram that's been cut in half, we can combine the logic of the previous two paragraphs to show that the quantity R·Nds is precisely the area of the parallelogram swept out between R and Tds, which is twice the area of the triangles formed when sweeping R around the curve. This logic generalizes to any dimension, as the dot product between a vector R and a normal vector N always represents the (hyper)volume swept out between a (hyper)area orthogonal to N and the vector R. This can also be phrased in terms of determinants if you're more comfortable with that language. For a visual demonstration of this idea search for the short video "Finding the Area of Arbitrary Polygons With Geometric Algebra" by Sudgylacmoe or alternatively look up the Shoelace Formula

[@caiollvllal]

I think he wanted a proof integrating by parts in the same scheme as the previous beauties. Idk about geometric algebra but your proof seems pretty too!

[@APaleDot]

@@caiollvllal I figured he would be satisfied with a proof that doesn't require the Divergence Theorem or introducing any kind of coordinate system.

[@burnytech]

so much love for math

[@connorshea9085]

It's important to specify this only works on a closed curve in 2D, where there is a single well-defined exterior normal direction, and where there is one defined area enclosed by the loop.

[@MathTheBeautiful]

Correct. Also works for surfaces in 3D.

[@Pluralist]

[@lih3391]

Have you heard of the generalized stokes theorem? It's definitely the most beautiful math theorem I've seen, and the divergence theorem is just one of it's applications. But the generalized stokes theorem extends that to any dimensional surface in any number of dimensions. It depends on no particular coordinate system, although non-cartesian coordinates make it much harder (tensor calculus). It's intuition is that the sum of local changes equals the global change. All the strange notation and definitions are to make this intuition a reality(not actually but it's still super useful in physics). From it, you get the fundamental theorem of calculus, divergence theorem, stokes theorem, etc.