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Given $z=f(x)$ for $a<x<b$ differentiable and $f(x) > 0$ for all $x\in (a,b)$. By revolution of the graph around the x-axis in $\mathbb{R}^{3}$ there evolves a revolutionary area $S$.

We parametrize: $\phi(s,t)= s, f(s) \sin(t), f(s) \cos(t)$

This gives: $$D\phi = \begin{pmatrix}1 & 0 \\ f' \sin(t) & f \cos(t) \\ f' \cos (t) & -f \sin(t) \end{pmatrix}$$ So : $$\|\partial_{1} \phi \times \partial_2 \phi \| = \sqrt{f^{2}f'+f^2 \sin^2 +f^2 \cos^2} = f\sqrt{1+f'(x)^2} \tag{$\ast$} $$

So now :$$\int_{0}^{2\pi} \int_{a}^{b} f \sqrt{1+f'(x)^2} dxdt = 2\pi \int_{a}^{b} f \sqrt{1+f'(x)^2} dx .$$

Is this last star step correct? Thanks for all efforts.

share|improve this question
    
Yes, insofar as the inner integrand is continuous on [0,2pi]. You might run into difficulties if $f'(x)$ has discontinuities here. I am not sure. –  01000100 Dec 16 '11 at 3:42
    
In the center square root on the line with * you should have $(f'(x))^2$. Then the last = will be correct. –  Ross Millikan Dec 16 '11 at 3:49
    
In the second formula of (*), you miss the square for the term $f'$. That is to say, $\|\partial\phi_1\times\partial_2\phi\|=\sqrt{f^2(f')^2+f^2\sin^2(t)+f^2\cos^2(t‌​)}$. –  Paul Dec 16 '11 at 3:53
1  
"Revolutionary area" should probably be "surface of revolution". –  Arturo Magidin Dec 16 '11 at 4:35

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