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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 $s$.

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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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Consider $ 2\pi \int_0^l \rho (s) \, ds $ We know $s(t)=\int_{a}^t||\alpha'(z)||dz$. For regular curves this is a diffeomorphism from $(a,b)$ to $(0,l)$ where $l$ is the arclength Thus using change of variable formula rewrite the integral as $2\pi\int_a^b\rho(s(t))|s'(t)|dt=2\pi\int_a^b|f(t)|\sqrt{f'(t)^2+g'(t)^2}dt$

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