# Surface area of revolution of trig fn

I want to fine the surface area of revoultion around the y-axis of $$x=cos^3(\theta), y=sin^3(\theta)$$

I looked up a answer below but it's confusing me. (note that it's revolution around x-axis)

To find the surface area, shouldn't it be $S=\int_{a}^{b} 2\pi y \;dx$ ? similar to disk method?

where does the $\sqrt{(dx)^2+(dy)^2}$ come from? I believe it has to do with line integral and have no idea what it does here.

If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is $$A=\pi(r_1+r_2)\ell\ ,$$ where $r_1$ and $r_2$ are the radii of the two ends and $\ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is $$d\ell=\sqrt{(dx)^2+(dy)^2} =\sqrt{\Bigl(\frac{dx}{d\theta}\Bigr)^2+\Bigl(\frac{dy}{d\theta}\Bigr)^2} d\theta\ .$$ So the area element is $$dA=2\pi f(x)\sqrt{\Bigl(\frac{dx}{d\theta}\Bigr)^2+\Bigl(\frac{dy}{d\theta}\Bigr)^2}d\theta\ .$$
• In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$. Jul 26, 2018 at 23:17