I have recently come across a problem that I think I understand but am unsure about.

I have been asked to integrate the Gaussian Curvature of the Torus onto the Torus itself. The parametrization is given as: $$T(\varphi, \theta) = ((R+r\cos(\theta))\cos(\varphi), (R+r\cos(\theta))\sin(\varphi), r\sin(\theta))$$

Now I computed the Geodesic Curvature of $\theta=\text{constant}, \varphi = \text{constant}$, which both gave $0$. I am told to integrate on three surfaces.

a) The Torus

b)The portion of the Torus bounded by $\theta\in[\frac{-\pi}{2}, \frac{\pi}{2}]$

c)The portion of the Torus bounded by $\varphi\in[\frac{-\pi}{2}, \frac{\pi}{2}]$

Now for a), I used the Gauss-Bonnet Theorem, assuming that $\kappa_G=0$ based on the Geodesic Curvatures giving zero. Then, I concluded:

$$\int_T K\cdot dA = 2\pi\chi(T)$$

The Euler Characteristic of the Torus is $0$, thus I concluded that the integral of $K\cdot dA=0$ on the torus. Now for b) and c), would I just integrate the $K$, which is: $$K= \frac{\cos(\theta)}{r(R+r\cos(\theta))}$$ as such?

b)$$\int_0 ^{2\pi} d\varphi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}K \sqrt{EG-F^2}d\theta = 0$$

c) $$\int_0^{2\pi} K \sqrt{EG-F^2}d\theta\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}d\varphi=0$$

I'm slightly confused as the integral gives $0$ in all three cases, I feel like I am going wrong somewhere. If someone could help it would be appreciated!


First of all, the circles $\phi=\text{constant}$ are in fact geodesics, but the circles $\theta=\text{constant}$ are only geodesics when $\theta=0$ or $\pi$. So your geodesic curvature computation is wrong. (Geometrically, the acceleration vector as you move at constant speed along the curves $\theta = \pm\pi/2$ is entirely tangent to the surface. Indeed, normal curvature is $0$ — these are asymptotic curves — and so all the curvature is geodesic curvature. One must check signs, however.)

If you check correctly, $\int K\,dA = 0$ when you integrate over the whole torus, as your Euler characteristic predicts. However, when you integrate over the outer "half" of the torus (part b), you should get a positive integral, not $0$ (note that $\cos\theta\ge 0$ when $\theta\in [-\pi/2,\pi/2]$). In this case, $\chi = 0$ and you do have negative geodesic curvature integrals. But pay attention to orientations on those curves! When you integrate over half the torus (part c), you do get integral $0$, the geodesic curvature integrals vanish, and $\chi = 0$, so no problem.

  • $\begingroup$ I'm not sure I have the right formula for geodesic curvature. Is it correct to say: $\endgroup$ – Felicio Grande Mar 28 '17 at 0:48
  • $\begingroup$ $$\kappa_G = \frac{T_{\theta\theta}\cdot(\vec n\times T_{\theta})}{<T_{\theta}, T_{\theta}>^{3/2}}$$ $\endgroup$ – Felicio Grande Mar 28 '17 at 0:50
  • $\begingroup$ $T$ should always denote the unit tangent vector (although earlier you used it for the torus). You can compute the curvature vector $\kappa N$ without an arclength parametrization by correcting with the chain rule. Take a look at my text, linked in my profile. $\endgroup$ – Ted Shifrin Mar 28 '17 at 0:54
  • $\begingroup$ Yes, the $T$ denoted here is for the torus. I had a look at your text, which I believe says that the Geodesic Curvature is given by:$$\kappa_G = \kappa N\cdot(\vec n\times T)$$ where I believe $T, n$ denote unit tangent, normal vectors respectively, and $N$ is the Frenet normal vector? $\endgroup$ – Felicio Grande Mar 28 '17 at 1:13
  • 1
    $\begingroup$ @FelicioGrande: In b) the integral of Gaussian curvature is $4\pi$, and the two geodesic curvature integral are each $-2\pi$ (be careful with orientations on the two circles). In this case $\chi=0$ and it comes out perfectly. $\endgroup$ – Ted Shifrin Mar 30 '17 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.