# Surface integral with differential forms

by gts87
Tags: differential, forms, integral, surface
 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 $$\phi(r,\theta) = (rcos\theta, rsin\theta, \sqrt{1 - r^2}), 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$$. 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: $$\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|\partial\phi/\partial r \times \partial\phi/\partial\theta| dr d\theta$$ $$\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|| dr d\theta$$ $$\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^4cos^2\theta + r^4sin^2\theta)/(1 - r^2) + r^2} dr d\theta$$ $$\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^2/(1 - r^2)} dr d\theta$$ $$\int^{2\pi}_{0}d\theta\int^{1}_{0}r\sqrt{1 - r^2} dr = 2\pi/3$$ However, using differential forms, if we let $$\omega = z^2 dx \wedge dy$$ 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!
 P: 1 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.

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