Question

Let \[ \Phi(u,v) = (u-v,u+v,uv) \] and let D be the unit disk in the uv plane. Find the area of \[\ \Phi(D)\] I got the right result by finding the \[\ ||\Phi_u\times \Phi_v || \] and then converting to polar coordinates. and integrating \[\ \int\int ||\Phi_r\times \Phi_\theta || rdrd\theta \] My question is: Would I get the same result if I first parametrized u and v as \[\ u = rcos\theta; v=rsin\theta\]So that \[ \Phi(r,\theta) = (rcos\theta-rsin\theta,rcos\theta+rsin\theta,r^2cos\theta sin\theta) \]and then computed \[\ ||\Phi_r\times \Phi_\theta || \], and would I have to add Jacobian?

Answer 1

I know I already asked about this, but I didn't get an answer. So I'm hoping I'll have more luck this time :D

Answer 2

Since you are just performing a change of basis from the u-v plane to the r-theta plane, this is perfectly allowed AS LONG AS you add the Jacobian of your transformation to the integral. Think of the Jacobian as a sort of correction factor in the new basis.

Answer 3

Ah, I thought that if I first parametrize it and then compute the normal that i would get that r from the cross product somehow. I thought that the Jacobian was somehow hidden in there and that I would get \[\ || \Phi_\theta \times \Phi_r|| = r*\sqrt{something}\] which would then give me \[\ \int\int \sqrt{something}* rdrd\theta \]

Answer 4

If for example I take \[\ \Phi(u,v) =(u,v,0) \] in the same unit disk I get \[\ \Phi(r,\theta) =(rcos\theta,rsin\theta,0) \]
\[\| \Phi_r\times\Phi_\theta\| = r \]
So, \[\ \int_{0}^{2\pi}\int_{0}^{1} r drd\theta = \pi r^2\] Which is correct, and I didn't add Jacobian. If I added it I would get r^2 under the integral. What am I thinking wrong here?