# Ever heard of Green's Theorem?

Jirakon
Posers and Puzzles 04 Jun '07 17:16
1. 04 Jun '07 17:16
Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

I'll use S for the integral sign.

Parameterize: x(t) = cos(t), y(t) = sin(t), 0
2. TheMaster37
Kupikupopo!
04 Jun '07 21:58
Originally posted by Jirakon
Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

I'll use S for the integral sign.

Parameterize: x(t) = cos(t), y(t) = sin(t), 0
So P = -sin(t) and Q = cos(t)

Then what?
3. 04 Jun '07 22:101 edit
Wait...what? I had just typed a whole proof of how 2pi = 0, but I used a less than sign after that 0 in the first post, which was probably misinterpreted as an incomplete tag. Let me try again:

Let P = -y/(x^2 + y^2), Q = x/(x^2 + y^2), and D be the unit disk.

I'll use S for the integral sign.

Parameterize: x(t) = cos(t), y(t) = sin(t), t is between 0 and 2pi inclusive

S Pdx + Qdy
S -sin(t) x -sin(t) + cos(t) x cos(t) dt
S sin^2(t) + cos^2(t) dt
S dt from 0 to 2pi = 2pi

SS dQ/dx - dP/dy dx dy
SS (y^2 - x^2)/(x^2 + y^2)^2 - (y^2 - x^2)/(x^2 + y^2)^2 dx dy
SS 0 dx dy = 0

Therefore 2pi = 0