Parametrisation of this triangle

Question

I have been doing an exercise problem, and there is something unclear about it. Namely, how was the triangle in the following example transformed to a circle $C'$, which is parametrised by $(x(t), y(t)) = (cost,sin t), 0 ≤ t ≤ 2π$?

Also, has the Green theorem actually been used here, since there is no double integral?

Answer 1

Green's theorem says that if the partial derivatives of $P,Q$ are defined everywhere in the region:

$\oint P\ dx + Q\ dy = \iint (\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}) dV$

Here you have the case that $\frac {\partial Q}{\partial x},\frac {\partial P}{\partial y}$ are not defined at the origin. But they are defined everywhere else.

Suppose divide the region into pieces.

$\oint_A P\ dx + Q\ dy + \oint_B P\ dx + Q\ dy + \oint_C P\ dx + Q\ dy$

Since we travel the contours both directions on the red segments, those are going to cancel each other out. The sum of the three contours integrals is the same as the contour integral around the triangle.

The partials are defined everywhere in regions A and B. We can use Green's theorem to evaluate those.

Since $(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}) = 0$ in those regions, $\oint_A P\ dx + Q\ dy + \oint_B P\ dx + Q\ dy = \iint 0 \ dA + \iint 0 \ dB = 0$

Which is a long way of getting to the point that all contours that enclose the origin (in this example) will have the same contour integral.

Answer 2

Walk around the triangle $C$ clockwise, move in to the unit circle along a line segment $S$, walk around $C'$ counterclockwise, and walk back to the triangle along $S$. You now have a curve entirely in the plane-minus-the-origin, and it bounds a kind of annular shape. Green's theorem, applied on that sliced-annulus, says that the sum of four integrals (over $C$, over $S$, over $C'$ backwards, and over $S$ backwards) sum to the integral of the divergence of $F$ on the interior, which is zero.

The two $S$ integrals cancel, and we conclude that the integral over $C$ and over $C'$ must be the same (so that summing the first with the negative of the second, because of the direction-reversal, gives zero).