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Let $S$ be a surface of revolution in $\mathbb{R}^3$ (2-dimensional) and let $C$ be its generating curve. Let $s$ be its arc lenght. Let $x = x(s)$ be the distance from a point in the curve to the $Oz$ axis (C lies in the $xz$ plane and we rotate it around $Oz$).

In my DG book, it says that

$Area(S) = 2x \int_0^l \pi(s) ds $. Where $l$ is the lenght of the curve. Although it says nowhere what this $\pi$ function is.

I've been able to prove, using the theory formulated in the book that

$Area(S) = 2\pi \int_0^l x(s) ds $ where this $\pi$ is the regular 3.14... $\pi$

The proof looks pretty correct to me. Is it safe to assume that the book was printed wrong?

thanks in advance.

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    "Differential Geometry of curves and surfaces" by Do Carmo This one2012-06-14

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Yes, it was typo. For the correct statement, you can look at p.101, Exercise 11 in Section 2-5 in the book "Differential Geometry of curves and surfaces" by Do Carmo.

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    Love that book, man. This is the book I got the typo from. maybe they corrected it in a more recent edition2012-06-14
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As remarked by others, it is a typo. The correct formula is

${\rm area}(S)=2\pi\int_0^l\ x(s)\ ds\ .$

Intuitively one can interpret this formula as follows: The surface $S$ is a union of "infinitesimal lamp shades" of radius $x(s)$ and width $ds$. The surface of such an "infinitesimal lamp shade" is the same as the surface of a plane circular annulus of radius $x(s)$ and width $ds$, namely $2\pi\ x(s)\ ds$.