Surface area of parametric curve $x = \cos^{3} \theta$, $y= \sin^{3} \theta$ rotating around $x$-axis

Question

Find the area of the surface obtained by rotating the curve about the $x - axis$.

$$x = \cos^{3} \theta$$ $$y = \sin^{3} \theta$$ $$0 < \theta < 2\pi$$

I've searched all around the internet and found:

But this quite doesn't make sense to me and neither does give me the correct answer as 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?

Answer 1

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 $$ S = \int_0^{2\pi} dt \int_0^\pi d\phi \left|\frac{\partial\vec r}{\partial t}\times \frac{\partial\vec r}{\partial \phi}\right|$$

Answer 2

The surface integral is

\begin{align} A &= 2\int_0^1 2\pi y\sqrt{1+(y’_x)^2}dx\\ &= 4\pi \int_{\pi/2}^0 \sin^3\theta \sqrt{1+\tan^2\theta}(-3\cos^2\theta\sin\theta)d\theta\\ &= 12\pi \int_0^{\pi/2}\sin^4\theta \cos\theta d\theta= \frac{12\pi}5 \end{align}