Theorem of Pappus

differential-geometrysurface-integralssurfaces

Given a surface of revolution $S$ which can be parametrized by the map
$$
\mathbf x(u,v) = (f(v)\cos u,f(v)\sin u,g(v)),
$$

over the open set $U =\{(u,v) \in \mathbb R^2 \mid 0 < u < 2\pi, a < v < b\}$, I computed the area of $S$ to be
\begin{align*}
\int_a^b\int_0^{2\pi} |\mathbf x_u \times \mathbf x_v| \, du \, dv = 2\pi\int_a^b |f(v)| \sqrt{(f'(v))^2+(g'(v))^2} \, dv.
\end{align*}

If $l$ is the length of the generating curve $C$, how does one then get the area of $S$ to also be written
$$
2\pi \int_0^l \rho (s) \, ds,
$$

where $\rho=\rho(s)$ is the distance to the rotation axis of the point $C$ corresponding to $s$? I think that the arc length $s=\int_a^b |\alpha'(t)| \, dt$, where $\alpha$ is the space curve, but I'm not sure in particular how one changes the interval $[a,b]$ to $[0,l]$ when changing the variable $v$ to $s$.

Best Answer

You need to reparametrize your curve, replacing $v$ by $s$, where $$ s(v)=\int\limits_a^v\sqrt{f'(v)^2+g'(v)^2}\,dv $$ hence $ds=\sqrt{f'(v)^2+g'(v)^2}\,dv$. When $v$ takes values in $[a,b]$, $s$ takes values in $[0,l]$ where $l$ is the length of the curve. Also, $|f(v)|$ is just the distance from the point to the rotation axis, so it is precisely $\rho(s)$ (you can do the computation, but I do not think there is a need for it),

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