# Parametrisation of the surface a torus

For a calculus question I need to parameterise the surface of the torus generated by rotating the circle given by $(x-b)^2+z^2=a^2$ around the $z$-axis (with $0<a<b$). I've had a go at this, and have come up with $$\vec{r}(\theta,\phi)=((a\cos\theta+b)\cos\phi, (a\cos\theta+b)\sin\phi, a\sin\theta), ~~ \theta,\phi\in [0,2\pi].$$

I haven't done anything like this for a while now, so I've kind of forgotten how it all goes. So I've tried plotting this on WolframAlpha to see if it's correct, but I can't get it to actually plot a surface defined in this way. I have a copy of Matlab, is there any way I could plot this surface on there to see if it's correct? Basically how can I check if I'm right or not? Failing that, could somebody please just point out if I'm wrong?

• as a British English speaker I'm pretty sure 'parameterise' is fine, but thanks for the edit anyway :) – Tim Apr 12 '13 at 15:34

Your formula is absolutely correct, just look here for verification.

• cannot believe I didn't think to check Wikipedia, sorry! but thank you very much – Tim Apr 11 '13 at 23:21
• @Tim no problem. – Coffee_Table Apr 11 '13 at 23:29

A circle of radius $a$ centered at $(b,0)$ in the plane $xz$ has the parametric equation

$$x=a\cos(\theta)+b,z=a\sin(\theta),$$ with $\theta$ in the range $[0,2\pi]$ for a full circle.

Now you rotate the plane $xz$ around $z$ by $x\leftarrow x\cos(\phi),y\leftarrow x\sin(\phi)$, with $\phi$ in the range $[0,2\pi]$ for a full turn,

$$x=(a\cos(\theta)+b)\cos(\phi),\\ y=(a\cos(\theta)+b)\sin(\phi),\\ z=a\sin(\theta).$$

If you freeze $\theta$, you get a circle in a plane parallel to $xy$, of the form:

$$x=r\cos(\phi),y=r\sin(\phi).$$

You can plot this surface in WolframAlpha. The correct syntax to use is:

ParametricPlot3D[{(2 Cos[u]+3) Cos[v], (2 Cos[u]+3) Sin[v], 2 Sin[u]}, {u, 0, 2 Pi}, {v, 0, 2 Pi}]


where I have chosen $a = 2$, $b = 3$. The output looks like this.