Integral of Gaussian Curvature over a Tube

Question

I am working on a problem which I have come to a standstill on. The question reads:

Let $\alpha(s)$ be a closed regular curve in $\mathbb{R}^3$ parametrized by arc-length. Consider the tube: $$X(s,v)=\alpha(s)+r(n(s)\cos(v)+b(s)\sin(v))$$ Where $n(s)$ and $b(s)$ represent the normal and binormal vectors of $\alpha(s)$. Integrate the Gaussain Curvature over the Tube.

Now I found the Gaussian Curvature (which was very tedious) and got the answer: $$K(s,v) = -\frac{\kappa(s)\cos(v)}{r(1-r\kappa(s)\cos(v))}$$ Where $\kappa(s)$ denotes the curvature of $\alpha(s)$. Now, how am I supposed to Integrate this over anything? I'm assuming that I should use Stokes', as it is the only thing that makes sense to do. So should I just consider $dK(s,v)$? and then consider the tube $T=\partial M$ as the boundary of a solid Tube $M$?

Answer 1

Use Gauss-Bonnet:

\begin{align*} \int_M KdA = 2\pi\chi(M). \end{align*}

Now the surface you discribe (somthing like a cylinder or torus) we have that $\chi(M)=0$. Thus $\int_M KdA =0$.