Let $T$ be a torus. We have a parameterization by $((c+a \cdot cos(v))cos(u),(c+a\cdot cos(v),a\cdot sin(v))$ for $u,v \in [0,2\pi)$. The first fundamental form is given by $E=(c+a\cdot cos(v))^{2}, F=0, G=a^2$

and the second fundamental form is given by $e=-(c+a\cdot cos(v))cos(v),f=0,g=-a$. Since gaussian curvature is equal to determinante of the 2. fundamental form divided by 1. first fundamental, the gaussian curvature is equal to $\frac{cos(v)\cdot a }{a^2(c+a\cdot cos(v))}$

Now I want to integrate the gaussian curvature over T.

Do I integrate the gaussian curvature over T correctly: $\int_{[0,2\pi)^2}\frac{cos(v)\cdot a }{a^2(c+a\cdot cos(v))} du dv$?

This integral seems to me quite uncomputable.

edit: I computed the gaussian curvature (probably) correctly, I found it on another site (see page 2,

  • 2
    $\begingroup$ Did you ever encounter the theorem of Gauß-Bonnet? Or are you trying to verify it? $\endgroup$ – Thomas Dec 10 '14 at 19:30
  • $\begingroup$ @Thomas: Yes, I'm aware of theorem of Gauss-Bonnet, which implies that this integral is zero. But I wanted to check it by computing the integral. I found the gaussian curvature of the torus on another site (see my post). Do I integrate correctly? $\endgroup$ – bjn Dec 10 '14 at 20:45

Let $K$ be the Gauss curvature. You would like to compute $\iint_{T} K \: dA$. Please remember that

\begin{equation*} dA = \sqrt{EG-F^2} du \: dv = a (c+a\cos v)\: du \:dv. \end{equation*}

Since you have already computed $K = \frac{a\cos v}{a^2(c+a\cos v)}$,

\begin{equation*} \iint_{T} K dA = \int_{0}^{2\pi} \int_{0}^{2\pi} \frac{a\cos v}{a^2(c+a\cos v)} \cdot a (c+a\cos v) \:du \: dv \end{equation*}

EDIT: Added some more detail about computing surface integrals.

  • $\begingroup$ I don't see my computation mistake. What's your first and second fundamental form? My computation shows that 1. first fundamental form is $$A:= \begin{pmatrix} a^{2} & 0 \\ 0 & (c+a \cdot cos(v))^{2} \end{pmatrix}$$. And my 2. fundamental form is $$B:=\begin{pmatrix} -(c+a \cdot cos(v))cos(v) & 0 \\ 0 & -a\end{pmatrix}$$. So the gaussian curvature is given by $\frac{det(B)}{det(A)}=\frac{a(c+a \cdot cos(v))cos(v)}{a^{2}(c+a \cdot cos(v))^{2} }$ $\endgroup$ – bjn Dec 11 '14 at 15:40
  • $\begingroup$ Thanks for additional computation steps. Now I see the problem: I do really integrate in the wrong way. I always struggle with this "informal" $dx$-Notation (see also my other post) in analysis (mostly in integration theory and differential equations). Is it a short informal way of doing a change of variables? What's exactly $dA$? I thought it's just some formal variable with no meaning and all that matters is, where we integate (in our case over $[0,2 \pi)^2$). Feel free to give me some references. $\endgroup$ – bjn Dec 11 '14 at 21:09
  • $\begingroup$ If you have any adives regarding the $dx$-notation, you can give an answer to this question math.stackexchange.com/questions/1035066/… $\endgroup$ – bjn Dec 12 '14 at 19:29

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