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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.

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What have you tried so far? If this is homework, use the (homework) tag. –  Loki Clock Mar 4 '13 at 20:42
Did you find an explicit definition, or is that what you need to know? In this sense of area-preserving? mathworld.wolfram.com/Area-PreservingMap.html –  Loki Clock Mar 4 '13 at 21:11
I think the definition of area-preserving that I was using was that the determinant of the metric was 1?? –  user55225 Mar 4 '13 at 21:21
Then see en.wikipedia.org/wiki/Metric_tensor#Area. –  Loki Clock Mar 4 '13 at 21:30
The same question as here math.stackexchange.com/questions/311789 –  Yuri Vyatkin Mar 5 '13 at 3:48
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1 Answer

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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)

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Thank you. Do you know how you would explicitly formulate the map $\varphi$ between the sphere and the cylinder? –  user55225 Mar 4 '13 at 21:26
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 4 '13 at 23:09
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