Surface area of ​the figure of rotation

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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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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

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