Surface integral with differential forms

[gts87]

Hi, I'm trying to solve a problem in David Bachman's Geometric Approach to Differential Forms (teaching myself.) The problem is to integrate the scalar function f(x,y,z) = z^2 over the top half of the unit sphere centered at the origin, parameterized by \phi(r,\theta) = (rcos\theta, rsin\theta, \sqrt{1 - r^2}), 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi. I think we can evaluate this surface integral using the formula \int\int_{S}f(x,y,z)dS = \int\int_{D}f(\phi(r, \theta))|\phi_{r}\times\phi_{\theta}|drd\theta yielding:

<br /> \int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|\partial\phi/\partial r \times \partial\phi/\partial\theta| dr d\theta<br />

<br /> \int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|&lt;r^2cos\theta/\sqrt{1 - r^2}, r^2sin\theta/\sqrt{1 - r^2}, r&gt;| dr d\theta<br />


<br /> \int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^4cos^2\theta + r^4sin^2\theta)/(1 - r^2) + r^2} dr d\theta<br />


<br /> \int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^2/(1 - r^2)} dr d\theta<br />


<br /> \int^{2\pi}_{0}d\theta\int^{1}_{0}r\sqrt{1 - r^2} dr = 2\pi/3<br />

However, using differential forms, if we let \omega = z^2 dx \wedge dy and use the same parameterization to integrate \omega over the mentioned manifold, we get

\int_{M}\omega = \int_{D}(1 - r^2)\cdot(\partial\phi/\partial r, \partial\phi/\partial\theta)dx \wedge dy (Here \partial\phi/\partial r, \partial\phi/\partial\theta are the tangent vectors being acted on by dx \wedge dy)


\int_{D}(1 - r^2) det[cos\theta \; -rsin\theta , \; \; sin\theta \; rcos\theta]drd\theta (the matrix rows are separated by the comma.)

\int^{2\pi}_{0}\int^{1}_{0}(1 - r^2)(r)drd\theta = \pi/2

Am I doing something wrong? If anyone can help I'd appreciate it, thanks!

 

[mirceap]

Hi!
In the first case you are doing nothing more than integrating the form f(\phi(r,\theta))\:\phi^*[dr\wedge d\theta]=f(\phi(r,\theta)) dS acting on the vectors \partial_r\phi,\:\partial_{\theta}\phi (here \phi^{*} represents the pushforward of a form). This corresponds to projecting the tangent vectors \partial_r\phi,\:\partial_{\theta}\phi to the plane (x,y) and then integrating on the disk 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi, with the factor |\partial\phi/\partial r \times \partial\phi/\partial\theta| accounting for the deformation due to projection to the horizontal plane. So this is the right result.

In the second case you are integrating f\;dx\wedge dy, but dx\wedge dy is not the surface element dS. Thus the result is different.