Find the area of the surface obtained by rotating the curve about the x - axis. x = cos^3 theta, y = sin^3 theta. 0 < theta < 2pi.

I've searched all around the internet and found: enter image description here

But this quite doesn't make sense to me and neither does give me the correct answer as enter image description here when rotated about x-axis, this part will not be counted as the surface area when multipled by two. So, how could I solve this question?

  • $\begingroup$ I don't know what tis black-red scribble is supposed to demark, and what is 'this part' that you're speaking of. But the formula you found is correct. You can also take the limits to be $t\in[0,\pi]$, and not multiply by 2, result is the same. $\endgroup$ – Adam Latosiński May 7 '19 at 13:58

The formula you found is correct, I don't understand what problem you have with it.

There's also another method:

You can parametrize the surface obtained with the rotation by $$ \vec{r}(t,\phi) = (x(t,\phi),y(t,\phi),z(t,\phi)) = (\cos^3 t,\sin^3 t\cos\phi,\sin^3 t\sin\phi)$$ with $t\in[0,2\pi]$, $\phi\in[0,\pi]$ (because you only need half of the rotation to get the whole surface). The area is then given by $$ S = \int_0^{2\pi} dt \int_0^\pi d\phi \left|\frac{\partial\vec r}{\partial t}\times \frac{\partial\vec r}{\partial \phi}\right|$$


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