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The coordinate patch of the surface $x(t, \theta) = (r(t)\cos\theta, r(t)\sin\theta, z(t))$ doesn't cover the entire surface of revolution, ommitting points that would correspond to $\theta = \pm\pi$.

How can we define a second coordinate patch with $0 < \phi < 2\pi$, check the overlap condition, and thus show that a surface of revolution is indeed a surface?

Thanks

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What are $r(t)$ and $z(t)$? –  Berci Sep 30 '12 at 8:36
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It is tacitly understood that a differentiable and regular meridian curve $$\mu:\quad t\mapsto \bigl(r(t),z(t)\bigr)\qquad(a<t<b)$$ with $r(t)>0$ is given. Then the map $$f:\ {\mathbb R}\times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\phi,t)\mapsto\bigl(r(t)\cos\phi,r(t)\sin\phi, z(t)\bigr)$$ has a differential $df$ of rank $2$ for all $(t,\phi)$, as $f_t\times f_\phi\ne 0$ everywhere. It follows that $f$ is an immersion, but it produces each piece of the intended surface of revolution $S$ an infinite number of times.

In order to obtain $S$ just once one could proceed to a quotient space ($\phi$ runs over ${\mathbb R}/(2\pi)$ instead of ${\mathbb R}$), or one covers $S$ with a finite number of "patches" that may overlap, but are related via "local coordinate transformations".

With only two patches there is the difficulty that they have a disconnected overlap, and two different local coordinate transformations apply in the two parts. Therefore I would propose to set up three patches as follows: $$p_1:\quad \bigl]{-{\pi\over2}},{\pi\over2}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\phi,t)\mapsto\bigl(r(t)\cos\phi,r(t)\sin\phi, z(t)\bigr)\ ,$$ $$p_2:\quad \bigl]{{\pi\over4}},{5\pi\over4}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\psi,t)\mapsto\bigl(r(t)\cos\psi,r(t)\sin\psi, z(t)\bigr)\ ,$$ $$p_3:\quad \bigl]{{3\pi\over4}},{7\pi\over4}\bigr[\ \times\ ]a,b[\ \to{\mathbb R}^3,\qquad (\omega,t)\mapsto\bigl(r(t)\cos\omega,r(t)\sin \omega, z(t)\bigr)\ .$$ In this way we have the three local coordinate transformations (only the angle variable is affected) $$p_1^{-1}\circ p_2: \quad \psi\mapsto \phi=\psi\qquad \bigl({\pi\over4}<\psi<{\pi\over2}\bigr)\ ,$$ $$p_2^{-1}\circ p_3: \quad \omega\mapsto \psi=\omega\qquad \bigl({3\pi\over4}<\omega<{5\pi\over4}\bigr)\ ,$$ $$p_3^{-1}\circ p_1: \quad \phi\mapsto \omega=\phi+2\pi\qquad \bigl({-{\pi\over2}}<\phi<{-{\pi\over4}}\bigr)\ .$$ Please note that everything we have written up here should be intuitively obvious and only served as an exercise in applying the general theory in a familiar case.

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how do we know that this surface of revolution is indeed a surface from the defined coordinate patches you suggested? Thanks –  mary Sep 30 '12 at 19:09
    
@mary: See my edit. –  Christian Blatter Sep 30 '12 at 19:50
    
Thanks for the help –  mary Sep 30 '12 at 23:20
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