Consider a tubular surface $S$, i.e. a surface which has a parametrisation
$$F:I\times\mathbb{R}\rightarrow \mathbb{R}^3,\quad F(t,\varphi)=c(t) + r · \Big(\cos \varphi · n(t) + \sin \varphi · b(t)\Big),$$
whereas $c:I\rightarrow\mathbb{R}^3$ is a regular curve with $||c'(t)||=1$ for all $t\in I $ and $\kappa(t)=||c''(t)||\ne 0$ for all $t\in I $.
I was able to show that the gauss curvature is
$$K=\frac{\det(h)}{\det(g)}=\frac{-r\big(1- r \cos(\varphi)κ(t)\big)\kappa(t)\cos(\varphi)}{r^2\big(1- r \cos(\varphi)κ(t)\big)^2}=-\frac{1}{r}\cdot\frac{\kappa(t)\cos(\varphi)}{\big(1- r \cos(\varphi)κ(t)\big)}$$
Now I want to show, that $$\int_S KdA=0$$
Since
$$\sqrt{\det(g)}=r\big(1- r \cos(\varphi)κ(t)\big)$$
I calculate
$$\int_S KdA=-\int_{I\times\mathbb{R}}\kappa(t)\cos(\varphi)\text{ }dt\text{ }d\varphi=-\int_{I\times\mathbb{R}}||c''(t)||\cos(\varphi)\text{ }dt\text{ }d\varphi,$$
But why is this equal to zero?
Do the iterated integral in the other order. What is $\displaystyle\int_0^{2\pi} \cos\phi\,d\phi$? (By the way, if you want a one-to-one parametrization of the tubular surface, you should not have $\phi\in\Bbb R$, of course.)