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Let $S$ be the complement of the points $(0,0,\pm{1})$ in $S^2$, and let $C=\{(x,y,z) \mid x^2+y^2 = 1,\ |z|< 1 \}$, be a cylinder of radius $1$. If $\varphi : S\to C$ is the map given by radial projection from the $z$-axis, show that $\varphi$ is area-preserving.

  • Did you find an explicit definition, or is that what you need to know? In this sense of area-preserving? http://mathworld.wolfram.com/Area-PreservingMap.html – Loki Clock Mar 04 '13 at 21:11
  • I think the definition of area-preserving that I was using was that the determinant of the metric was 1?? –  Mar 04 '13 at 21:21
  • Then see http://en.wikipedia.org/wiki/Metric_tensor#Area. – Loki Clock Mar 04 '13 at 21:30
  • Thank you, this is the definition that I am using, however, I do not know what map I need to construct between the two spaces. If I can figure this out, I should be able to work out $E,F$ and $G$ and then show that $\sqrt{EG-F^2}=1$. –  Mar 04 '13 at 21:32
  • Let $C(x,y)=\theta$ where $\theta=arccos(\frac{x}{x^2+y^2})$. Then let $P(\theta,z)=(sin(\theta),cos(\theta),z)$.

    Then $\phi=P(C(x,y),z)$.

     – Loki Clock Mar 04 '13 at 21:39
  • Doesn't this $\varphi$ depend on $x,y$ and $z$. I think that to be able to calculate E,F,G I need to find $d \varphi$, with $\varphi$ just a function of two variables. I think I may be getting confused between the parametrisation of surfaces and maps between surfaces. –  Mar 04 '13 at 21:42
  • http://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder – Anton Petrunin Mar 05 '13 at 03:41
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    The same question as here http://math.stackexchange.com/questions/311789/ – Yuri Vyatkin Mar 05 '13 at 03:48

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By rotational symmetry, it is enough to show that the area of an annulus (points with latitude betwee $\alpha$ and $\beta$) is porportional to its height $\sin\beta-\sin\alpha$. If $\alpha\approx\beta$, this can be approximated by the lateral surface of a chopped off cone, which again is proportional to the circumference $\approx 2\pi \cos\alpha$ times the tilted height $\beta-\alpha$. Ultimately this boils down to the fact that the derivative of $\sin$ is $\cos$ (or the integral of $\cos$ is $\sin$, depending on perspective)

Hagen von Eitzen
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  • Thank you. Do you know how you would explicitly formulate the map $\varphi$ between the sphere and the cylinder? –  Mar 04 '13 at 21:26
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    Sure, $(x,y,z)\mapsto\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}},z\right)=\left(\frac x{\sqrt{1-z^2}},\frac y{\sqrt{1-z^2}},z\right)$ – Hagen von Eitzen Mar 04 '13 at 23:09