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For example, I want to rotate a curve $z = x^3$ around the axis $OZ$. I have a surface: $z = x^3 + y^3 $. Now I want to compute a area of this surface.

Parametrize it:

$ x = rcos\phi \\ y=rsin\phi \\ z = r^3(cos^3\phi + sin^3\phi) $

Now how changes $r$ and $\phi$? We want compute a area of bounded surface. Let the plane $z=a$ cross our figure from below and $z=b$ from above.

Consequently, $r_{1} = (\frac{a}{cos^3\phi + sin^3\phi})^{1/3}; r_{2}=(\frac{b}{cos^3\phi + sin^3\phi})^{1/3} $ and $\phi \in [0, 2\pi] $.

Now I need to compute surface integral:

$ \iint_{S} 1\cdot dS = \int_{0}^{2\pi}d\phi \int_{r1}^{r2} \sqrt{EG - F^2}dr$, where $E = |\frac{\partial \overline p}{\partial r}|^2, G = |\frac{\partial \overline p}{\partial \phi}|^2, F=(\frac{\partial \overline p}{\partial r}, \frac{\partial \overline p}{\partial \phi})$ - scalar product, where $\overline p$ - parametrization.

And here I'm stuck. Because computing of $\sqrt{EG - F^2}$ is very hard. Then integration hard too. How I can do similar tasks easier? And how in the general case does it write down the equation of the surface of rotation?

Thank you in advance!

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    $\begingroup$ Refer to en.m.wikipedia.org/wiki/Surface_of_revolution $\endgroup$
    – user
    Commented May 19, 2018 at 11:13
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    $\begingroup$ Oh, thank you! In Russian Wikipedia, there is almost nothing. It is necessary to get used to reading foreign. $\endgroup$
    – mathmaniac
    Commented May 19, 2018 at 11:32

1 Answer 1

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HINT

A way to compute this kind of surface integrals is by the following set up

$$S=\int_a^b 2\pi f(z) \sqrt{1+[f’(z)]^2}\, dz$$

with

$$f(z)=\sqrt[3] z$$

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  • $\begingroup$ Why $f(z)=\sqrt[3] z$? Because $f$ depends on $z$ and from function we "make" another function: $x = \sqrt[3] z$ then write this integral? $\endgroup$
    – mathmaniac
    Commented May 19, 2018 at 11:34
  • $\begingroup$ And how can I derive this formula? $\endgroup$
    – mathmaniac
    Commented May 19, 2018 at 11:39
  • $\begingroup$ Yes of course, exactly! $\endgroup$
    – user
    Commented May 19, 2018 at 11:39
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    $\begingroup$ It’s simple $2\pi f(z)$ is the circumference, that is $f(z)$ is the radius. The other term derives by Pythagoras since the infinitesimal length to be considered is $\sqrt{dz^2+dx^2}$. Can you derive that? $\endgroup$
    – user
    Commented May 19, 2018 at 11:43
  • $\begingroup$ Yes, it seems understood. And last, if I had, for example, a curve $y = x^k$ where $k \in \mathbb{N}$ which rotate around axis $OY$. I have:$S=\int_a^b 2\pi f(y) \sqrt{1+[f’(y)]^2}\, dy $? Why we have in your hint $f(z)$, not $f(x)$? Thank you very much for your answers! $\endgroup$
    – mathmaniac
    Commented May 19, 2018 at 11:50

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