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I have a parametric representation of a curve: $$\gamma : [a,b] \rightarrow (r,z) \in \mathbb R^+ \times \mathbb R$$

and I want to find a general formula of the surface area of the revolution given by: $$\phi:(u,v) \in [a,b]\times[0,2\pi] \mapsto (\gamma_r(u)\cos v, \gamma_r(u) \sin v, \gamma_z(u))$$

Since I know the formula for general surfaces is:

$$A = \int_K \Big | \Big |\frac{\partial\phi}{\partial u} (u,v) \times\frac{\partial\phi}{\partial v }(u,v) \Big | \Big | d(u,v)$$

I started by calculating the normal vector, so:

$$\frac{\partial\phi}{\partial u} (u,v)= (\gamma_r(u)'\cos v, \gamma_r(u)' \sin v, \gamma_z(u)') $$ $$\frac{\partial\phi}{\partial v} (u,v) = (-\gamma_r(u)\sin v, \gamma_r(u) \cos v, 0) $$

and by taking the cross product I got $$\frac{\partial\phi}{\partial u} (u,v) \times\frac{\partial\phi}{\partial v }(u,v) = (\gamma_z(u)'\gamma_r(u)\cos v, \gamma_r(u)\sin v, - \gamma_r(u)\gamma_r(u)')$$

This is where I got stuck however... When taking the norm and then the integral, I cannot really simplify it a lot more. I think I have to substitute for $\gamma$, but I am pretty lost right now. I do not know how to get rid of the functions $\gamma_r$ and $\gamma_z$. Any help would be greatly appreciated!

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  • $\begingroup$ Do you want the formula for the "surface" or for the "area of the surface"? $\endgroup$ – user247327 May 20 '17 at 12:55
  • $\begingroup$ @user247327 I meant the area, will update the question $\endgroup$ – AxiomaticApproach May 20 '17 at 12:56
  • $\begingroup$ Not clear what exactly you want. You have written parametrization in your second equation. $\endgroup$ – Narasimham May 20 '17 at 12:57
  • $\begingroup$ @Narasimham I have the surface given by that parametrization only, and I want to find a general formula for the area of that surface $\endgroup$ – AxiomaticApproach May 20 '17 at 13:00
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That norm is $\sqrt{\gamma_r(\gamma'_r+ \gamma'_z)}$. That is a function of u only. The "general formula" for surface area is $2\pi\int_a^b\sqrt{\gamma_r(u)(\gamma'_r+ \gamma'_z)}du$. Of course, what that is, and whether it is expressible in a simple form, depends strongly on what $\gamma_r(u)$ and $\gamma_z(u)$ are.

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  • $\begingroup$ Aren't those $\gamma$'s squared? $\endgroup$ – Rafa Budría May 21 '17 at 21:07
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Easy to find $E,F,G$ of first fundamental form.Then,

$$ A =\sqrt{EG-F^2} du \, dv $$

From elementary differential calculus

$$ A = \int 2 \pi r \sqrt {dr^2 +dz^2}$$ for full $v.$

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  • $\begingroup$ Thanks! I have however never heard of the first fundamental form, so I think I will not be able to use this unfortunately.. $\endgroup$ – AxiomaticApproach May 20 '17 at 13:17
  • $\begingroup$ It is classical surface theory, time perhaps to look it up. $\endgroup$ – Narasimham May 21 '17 at 12:24

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