gts87
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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)|<r^2cos\theta/\sqrt{1 - r^2}, r^2sin\theta/\sqrt{1 - r^2}, r>| 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!
<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)|<r^2cos\theta/\sqrt{1 - r^2}, r^2sin\theta/\sqrt{1 - r^2}, r>| 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!