Flux through two parametrized surfaces

[frozenguy]

Homework Statement

Show that the flux through a parametrized surface does not depend on the choice of parametrization. Suppose that the surface $$\sigma$$ has two parametrizations, r(s,t) for (s,t) in the region R of st-space, and also r(u,v) for (u,v) in the region T of uv-space, and suppose that the two parametrizations are related by a change of variables: u = u(s,t), v = v(s,t). Suppose that the Jacobian determinant $$\frac{\partial(u,v)}{\partial(s,t)}$$ is positive at every point (s,t) in R Use the change of variables formula for double integrals to show that computing the flux integral $$\Phi=\int\int Fnds$$ parametrization gives the same result.

The Attempt at a Solution

So if the Jacobian determinant is positive at every point in R, what does that mean for T? That it is positive as well?
Have I used to the change of variables formula to show that computing the flux using either parametrization gives the same result?

Thanks for your help.

 

[kurt.physics]

Yes, the Jacobian determinant is positive in both regions R and T. The change of variables formula for double integrals states that if we have two different parametrizations of the same surface (which is what you have here), then the flux integral does not depend on the choice of parametrization. This is because the value of the flux integral is the same regardless of which parametrization is used. To show this, we can use the formula for double integrals: \Phi=\int\int Fnds = \int\int F(u,v)\frac{\partial(s,t)}{\partial(u,v)}dudv Since the Jacobian determinant is positive at each point in both regions, the two integrals will have the same value.