I need to compute the surface area of the torus $$T^2=\{(x,y,z)\subseteq\mathbb R^3 \left(\sqrt {x^2+y^2}- R\right)^2+z^2=r^2\}$$ where $0<r<R$.
I know I need to compute the metric tensor and the Gramian determinant etc, but in order to so, I need a regular global parametrization of $T^2$, I guess? How do obtain the latter for such a set?
Some help is appreciated.
On
Note you have a sum of squares equal a constant. Hence you should use trigonometric functions. $$ \left(\sqrt{x^2+y^2}-R\right)^2+z^2=r^2 $$ So we set $z=r\cos(\phi),\sqrt{x^2+y^2}-R=r\sin(\phi)$. Squaring the second relation we get $$ x^2+y^2=(R+r\sin(\phi))^2 $$ So we may introduce another angle an set $x=(R+r\sin(\phi))\cos(\theta),y=(R+r\sin(\phi))\sin(\theta)$. And this gives you the natural parametrization of the torus (You can see for ur self that those angles actually have a geometric meaning) $$ x=(R+r\cos(\phi))\cos(\theta)\\ y=(R+r\sin(\phi))\sin(\theta)\\ z=r\cos(\phi) $$
On
Here's an easy derivation of the surface area of a torus (without using metric tensor or the Gramian determinant).
All we have to do is compute the following integral:
$$2\pi\int_{0}^{2\pi}\left(R+r\cos\theta\right)\cdot rd\theta$$
Which can be explained through this diagram:
Since $\displaystyle \int(R+r\cos\theta)\cdot rd\theta=r\left(r\sin(\theta)+R\theta\right)+C$ , we conclude that
$$\displaystyle2\pi\int_{0}^{2\pi}\left(R+r\cos\theta\right)\cdot rd\theta=4\pi^2Rr$$
We can describe the surface of the torus using parameters $(\phi,\alpha)$ by the position vector $\vec r(\phi,\alpha)$
$$\begin{align} \vec r(\phi,\alpha)&=\hat \rho(\phi) R+(\hat \rho(\phi) r\cos(\alpha)+\hat z r\sin(\alpha))\\\\ &=\hat \rho(\phi)\,(R+r\cos(\alpha))+\hat z r\sin(\alpha) \end{align}$$
where $\hat \rho(\phi)=\hat x \cos(\phi)+\hat y \sin(\phi)$ is the radial unit vector in cylindrical coordinates, $0\le \phi<2\pi$, and $0\le \alpha<2\pi$.
Note that the angle $\alpha$ can be interpreted as the polar angle in a local cylindrical coordinate system centered at points on the axis of the torus.