I am trying to compute total curvature of $F(x,y) = x^4+y^4-1=0$.
My first trial)
I compute the curvature
\begin{align} \kappa = \frac{- F_y^2 F_{xx} + 2 F_x F_y F_{xy} - F_x^2 F_{yy}}{(F_x^2 + F_y^2)^{\frac{3}{2}}} = - \frac{3x^2 y^2 (x^4+y^4)}{(x^6+y^6)^{\frac{3}{2}}} \end{align} and then I tried to compute \begin{align} \int_{F(x,y)} \kappa ds \end{align} with $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}dt$
From $ F = x^4 + y^4 -1 \qquad \Rightarrow \quad \frac{dx}{dt} = - \frac{y^3}{x^3} \frac{dy}{dt} \quad \Rightarrow \quad ds = (\frac{x^6+y^6}{x^6})^{\frac{1}{2}} \frac{dy}{dt} dt $ I tried to compute $\kappa$ but its seems too complicated.
Actually I tried to compute this integral with $x=\cos^2(t)$ and $y=\sin^2(t)$ with $t\in [0,2\pi]$ via mathematica, but it didn't give answer.
My second trial) In this case I want to paramatrized first and use following formula \begin{align} &X(t) , \quad \kappa = \frac{|X'\times X''|}{|X'|^3}, \quad \frac{ds}{dt} = |X'| \\ &\int_{X} \kappa ds = \int_X \frac{|X'\times X''|}{|X'|^2} dt = \end{align} Then I realize that $X = (\pm \sin^{\frac{1}{2}}t, \pm \cos^{\frac{1}{2}} t)$, in this case I have trouble with finding proper region.
How one can compute the total curvature for $F(x,y) = x^4+y^4-1$?