some integral of curvature I have a problem :
Let $C$ be a curve defined by $C := \{\ (x,y)\in \mathbb{R}^2 |\ x^4 + y^4 = 1\}$,
and let $k$ be its curvature.
Compute the integral  $\int_C k$
This is an problem in a exam
What is the definition of the integral?
And how can I evaluate?
 A: For any curve in $\mathbb{R}^d$, let $s \mapsto \vec{x}(s)$ be a parametrization of the curve by its arc length. In terms of this parametrization, the tangent vector $\vec{t}(s)$, normal vector $\vec{n}(s)$ and curvature $\kappa(s)$ are "defined" by the relations:
$$\frac{d}{ds} \vec{x}(s) = \vec{t}(s) 
\quad\quad\text{ and }\quad\quad
\frac{d}{ds} \vec{t}(s) = \kappa(s) \vec{n}(s),\;\;|\vec{n}(s)| = 1
$$
Please note that in this definition, there is an ambiguity which direction the normal vectors $\vec{n}(s)$ point to. The common convention is pick the directions such that $\vec{n}(s)$ 
is as smooth as possible. If the curve doesn't have any point of inflection, i.e the curvature $\kappa(s)$ never vanishes, there are essentially only two "set" of choices
of directions of $\vec{n}(s)$ over the whole curve.
One consequence of above definition is
$$|\kappa(s)| = \left|\frac{d}{ds}\vec{t}(s)\right|\tag{*1}$$
For any curve on the plane, let $\theta(s)$ be the angle between $\vec{t}(s)$ and $x$-axis. 
Same as $\vec{n}(s)$, there is an ambiguity of what value of $\theta(s)$ should be. This is because $\theta(s)$ is well defined only up to a multiple of $2\pi$. The convention is
again choose $\theta(s)$ to be as smooth as possible over the whole curve.
In terms of $\theta(s)$, we have:
$$\vec{t}(s) = (\cos \theta(s), \sin \theta(s))\tag{*2}$$
Substitute $(*2)$ into $(*1)$, one find
$$|\kappa(s)| = \left|\frac{d\theta(s)}{ds}\right|$$
For the curve $x^4 + y^4 = 1$, if you walk around it in ccw direction, you will find
$\frac{d\theta(s)}{ds} > 0$ and the corresponding $\theta(s)$ increases for $2\pi$ after you return to original spot$\color{blue}{^{[1]}}$. This means up to a sign, the integral you want to evaluate is
$$\int \kappa(s)ds = \int \frac{d\theta(s)}{ds} ds = \int_0^{2\pi} d\theta = 2\pi$$
Notes
$\color{blue}{[1]}$ 
Let $L$ be the length of the curve, we wish to justify the assertion $\theta(L) - \theta(0) = 2\pi$ here.
Let $( r(s), \varphi(s) )$ be the polar coordinate of the curve. i.e.
$$s\quad\mapsto\quad\vec{x}(s) = ( r(s)\cos\varphi(s), r(s)\sin\varphi(s) )$$
WOLOG, we will also assume $s$ is chosen such that $\varphi(s) = 0$.
One observation is that when $\theta(s)$ are chosen appropriately, then $0 < \theta(s) - \varphi(s) < \pi$ over the whole curve. Since $\varphi(L) = 2\pi$, we get
$$| \theta(L) - \theta(0) - 2\pi | = |(\theta(L) - \varphi(L)) - (\theta(0) - \varphi(0))| < 2\pi\tag{*3}$$
Since $s = L$ and $s = 0$ represent the same point on the curve, $\theta(L) = \theta(0)$
up to an integer multiple of $2\pi$. $(*3)$ then implies the difference is exactly $2\pi$.
A: Use Global Gauss-Bonnet Theorem : 
$$ \int_{C = \partial R} k_g +\int\int_R K =2\pi \chi (R)\ ( \ast)$$
Since $R$ is a plane so $\chi(R)=1$ and Gauss curvature $K$ is $0$. 
Also, geodesic curvature $k_g$ is equal to curvature $k$ because of plane. 
Hence $2\pi$. 
( Given curve is just a simple closed curve with no singularity so that there exists no external angle term in $\ast$ )
