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Let $f:\mathbb R\to \mathbb R$ a smooth function. We suppose $f> 0$. Let $\gamma =\{(0,x,f(x))\mid x\in [0,1]\}$ and let $\Sigma$ the revolution surface obtained by the rotation of $\gamma $ around the axis $Oy$. Prove that $$Surface (\Sigma)=\int_0^1 2\pi f(x)\sqrt{1+f'(x)^2}dx,$$ using Riemann sum only. The idea is the following one : an infinitesimal element of area is the red "annulus around a cone". The area is given by $$\mathcal A(x+h)-\mathcal A(x)=2\pi f(x)\sqrt{1+f'(x)^2}h+o(h).$$

Let $\{x_i^n\}_{i=0}^{m_n-1}$ subdivision of $[0,1]$ s.t. $\Delta x_i^n:= x_{i+1}^n-x_i^n\to 0$ when $n\to \infty $. We have that

$$Surface (\Sigma)=\lim_{n\to \infty }\sum_{i=0}^{m_n-1}(\mathcal A(x_{i+1}^n)-\mathcal A(x_i^n)$$ $$=\lim_{n\to \infty }\sum_{i=0}^{m_n-1}2\pi f(x_{i}^n)\sqrt{1+f'(x_i^n)}\Delta x_i^n+\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)$$ $$=\int_0^1 2\pi f(x)\sqrt{1+f'(x)^2}dx+\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n).$$ How can I prove now that $\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0$ ?

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  • $\begingroup$ For such a proof we need the chosen setup for the "area" of curved surfaces. $\endgroup$ – Christian Blatter Aug 14 at 15:36
  • $\begingroup$ @ChristianBlatter: What do you mean ? Here it's normally not a complicated problem, is it ? It's juste the revolution surface obtained by rotating the curve around $Oy$. It's not true that $\lim_{n\to \infty }\sum_{i=0}^{m_n-1}o(\Delta x_i^n)=0$ ? If not, could you provide a counter example ? In all example I have in mind it works... $\endgroup$ – user657324 Aug 14 at 16:00

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