I was trying to calculate the surface area of a torus, whose tube radius is r, and distance from "singularity" to the center of the torus tube is R.
Here's what I've tried to do (The reason I think I'm wrong is because of wolfram alpha torus page).
I should note that the same methode did yield the right volume.
Best Answer
Use the fact that the torus is a surface of revolution. Consider a torus generated by the rotation of the circle $$(x-R)^2 + y^2 = r^2, \quad R > r > 0$$ about the $y$-axis. Then by symmetry, consider the top half given by $$f(x) = (r^2 - (x-R)^2)^{1/2}, \quad R-r \le x \le R+r.$$ Then in the interval $$x \in [x_0, x_0 + \Delta x],$$ the differential arc length of $f(x)$ at $x_0$ is $$\Delta s \approx \sqrt{(\Delta x)^2 + (f(x_0 + \Delta x) - f(x_0))^2} = \Delta x \sqrt{1 + \left(\frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}\right)^2},$$ and as $\Delta x \to 0$, we see that the squared term is a derivative; hence $$ds = \sqrt{1 + (f'(x))^2} \, dx$$ at each point of integration along the $x$-axis. This formula should be quite familiar. Then the differential surface area of this differential arc length as rotated about the $y$-axis is given by $$dA = 2 \pi x \, ds = 2\pi x \sqrt{1 + (f'(x))^2} \, dx.$$ Consequently, for half of the torus (the half above the $x$-axis), we find $$f'(x) = \frac{1}{2}(r^2 - (x-R)^2)^{-1/2} \cdot 2(R-x) = \frac{R-x}{f(x)},$$ and $$\sqrt{1 + (f'(x))^2} = \frac{r}{f(x)};$$ hence the total surface area is $$A = 2\pi r \int_{x=R-r}^{R+r} \frac{x}{f(x)} \, dx.$$ To make the calculation easier, we consider the substitution $$u = \frac{x-R}{r}, \quad dx = r \, du,$$ resulting in $$A = 2 \pi r \int_{u=-1}^1 \frac{ru \, du}{\sqrt{1-u^2}} + \frac{R \, du}{\sqrt{1-u^2}}.$$ The first term is clearly $0$, being the integral of an odd function over $[-1,1]$, leaving the second term: $$A = 2\pi Rr \int_{u=-1}^1 \frac{du}{\sqrt{1-u^2}} = 2\pi^2 Rr,$$ after a standard transformation $u = \sin \theta$. Therefore, the total surface area of the torus is $2A = 4\pi^2 Rr$.