Gaussian Curvature of $x^4+y^4+z^4=1$

Question

Let $S=\{(x,y,z)\in \mathbf R^3 | x^4+y^4+z^4=1 \}$ .
To compute the Gaussian curvature $k$ of $S$, I tried an elementary method to find $dN_p$. Let $\alpha (t) = (x(t),y(t),z(t))$ be an parametried curve on $S$. Since $4x^3x'+4y^3y'+4z^3z'=0$, we obtain the normal map $N(t)=\dfrac{1}{n(t)}(x^3(t),y^3(t),z^3(t))$ where $n(t)$ is the absolute value of the vector $(x^3(t),y^3(t),z^3(t))$. Then we can find the explicit form of $dN_P$ from the relation $dN_p(\alpha ' (0))=N'(0)$.
However $N'(t)$ is much complicated by $n(t)$. So I doubt whether my approach is right. Is there an easy method of computing $k$?
My ultimate goal is to compute $\displaystyle \int_S k$. Should I try another approach?

Answer 1

Using the nice formula at wikipedia. I calculate: $$ K = \frac{9x^2y^2z^2}{x^6+y^6+z^6}. $$

Answer 2

You can use the shape operator. You know that $U = \frac{1}{4\sqrt{x^6 + y^6 + z^6}}(4x^3,4y^3,4z^3)$ is the unit normal vector to this object. Now the shape operator is defined by $$S_p(v) = - \nabla_vU.$$ The gaussian curvature is defined by $$k_g := \det S_p.$$