# Parametric Hacks (I have the solution but something has gone wrong)

1. May 8, 2009

### Quantumpencil

1. The problem statement, all variables and given/known data Check Stokes Theorem for the Vector Field F(x,y,z)=(z,x,y), on the region defined by z=x^2-y^2 and x^2+y^2<=1.

Attempt at Solution:

Should be pretty easy problem; I tried two parameterizations, one worked and the other didn't... but I feel like they are equivalent. They were as follows: The boundary of the surface, defined when x^2+y^2 = 1, can be given by the paremetrization:

r(t)=(t, Sqrt[1-t^2], 2t^2-1)
r(t)=(Cos [Theta], Sin[Theta], Cos[2Theta]

Then the line integral along this path Should just be... the integral dt, from 0 to 1, or the integral dTheta from 0 to 2Pi of the dot product of F(r(t)) and r'(t), where r'(t) is given by

r'(t)=(1, -t/Sqrt[1-t^2], 4t)
r'(t)=(-Sin[Theta], Cos[Theta], -2Sin[2Theta]

Now, using the second parameterization, I got the value of this integral to be Pi. This is the right Answer. Using the first, I got the answer 1-Pi/4... but aren't they exactly equivalent?

I also evaluated the Surface integral of the Curls ((Curl F = 1,1,1)), and there got Pi using both co-ordinate systems (The integrals running from -Sqrt[1-x^2], to Sqrt[1-x^2] dy and -1 t0 1 dx, and from -1 to 1 dr and 0 to 2Pi dTheta, respectively)

Could someone help me figure out what is wrong with my first parameterization?

2. May 8, 2009

### HallsofIvy

Staff Emeritus
I didn't work this out in detail, but one thing I notice is that $y= \sqrt{1- x^2}$ only gives half of boundary: the half with y> 0. The two parameterizations are NOT equivalent.