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Surface integral with differential forms

by gts87
Tags: differential, forms, integral, surface
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Jan11-09, 08:04 PM
P: 6
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 [tex]\phi(r,\theta) = (rcos\theta, rsin\theta, \sqrt{1 - r^2}), 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi[/tex]. I think we can evaluate this surface integral using the formula [tex]\int\int_{S}f(x,y,z)dS = \int\int_{D}f(\phi(r, \theta))|\phi_{r}\times\phi_{\theta}|drd\theta[/tex] yielding:

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

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

\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

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

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

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

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

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

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

Am I doing something wrong? If anyone can help I'd appreciate it, thanks!
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Jan12-09, 02:31 AM
P: 1
In the first case you are doing nothing more than integrating the form [tex]f(\phi(r,\theta))\:\phi^*[dr\wedge d\theta]=f(\phi(r,\theta)) dS[/tex] acting on the vectors [tex]\partial_r\phi,\:\partial_{\theta}\phi[/tex] (here [tex] \phi^{*}[/tex] represents the pushforward of a form). This corresponds to projecting the tangent vectors [tex]\partial_r\phi,\:\partial_{\theta}\phi[/tex] to the plane (x,y) and then integrating on the disk [tex]0 \leq r \leq 1, 0 \leq \theta \leq 2\pi[/tex], with the factor [tex]|\partial\phi/\partial r \times \partial\phi/\partial\theta|[/tex] accounting for the deformation due to projection to the horizontal plane. So this is the right result.

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

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