# Integrating the Gaussian Curvature of the Torus on the Torus and Subsets of the Torus

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!

## 1 Answer

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.

• I'm not sure I have the right formula for geodesic curvature. Is it correct to say: – Felicio Grande Mar 28 '17 at 0:48
• $$\kappa_G = \frac{T_{\theta\theta}\cdot(\vec n\times T_{\theta})}{<T_{\theta}, T_{\theta}>^{3/2}}$$ – Felicio Grande Mar 28 '17 at 0:50
• $T$ 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. – Ted Shifrin Mar 28 '17 at 0:54
• 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? – Felicio Grande Mar 28 '17 at 1:13
• @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. – Ted Shifrin Mar 30 '17 at 1:08