Let $S \subset \mathbb{R}^3$ be the surface given by $x^4+y^4+z^4=1$, and let $K$ be its gaussian curvature. Then what is $\int_S K$?
First of all, I think finding patch seems hard. Should I consider 8 patches given by $(x,y,z)=(\pm \sqrt{\sin u \cos v},\pm \sqrt{\sin u \sin v},\pm \sqrt{cosu})$ where $0<u,v<\pi/2$? But then how should I deal with boudaries of the patches(for example, $u=0, v=0$)?
I guess you are expected to apply the Gauss-Bonnet theorem rather than calculate directly.
But if you really want to integrate $K$, the boundaries of patches can be ignored. The surface is smooth, hence there is no singular component to its curvature. That is, $K$ is a continuous function, and as such integrates to zero over any set of measure zero.
If the surface was $|x|+|y|+|z|=1$ instead, then you would have to worry about the boundaries between triangular faces: in fact, all of the curvature is concentrated there. (This is an octahedron, obviously non-smooth).
I changed the notation up a little bit. In the following, I call the surface defined by $x^4 + y^4 + z^4 = 1$ $Q^2$, for "two (dimensional) quartic", instead of $S$ as does the OP; done since I need the two-sphere $S^2$ below and I wanted to avoid confusion.
The key here is to realize that the surface $x^4 + y^4 + z^4 = 1$ is diffeomorphic to the $2$-sphere $S^2$, or more specifically, they each have Euler-Poincare characteristic equal to $2$. Let us denote the given surface by $Q^2$; to establish a specific diffeomophism $\phi:S^2 \to Q^2$, consider, for any point $(x, y, z) \in S^2$, the ray $t(x, y, z)$, $0 \le t < \infty$. I claim this ray intersects $Q^2$ for a unique value of $t \ge 1$, and that $t = 1$ precisely for the points $(x, y, z) \in S^2$ such that $(x, y, z) = (\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)$. This is most easily seen by observing that, if $t(x, y, z) \in Q^2$, then
$t^4(x^4 + y^4 + z^4) = 1 \tag{1}$
with
$x^2 + y^2 + z^2 = 1. \tag{2}$
Now if $(x, y, z) \in \{(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)\}$, (1) is clearly satisfied with $t = 1$; these points lie in $S^2 \cap Q^2$. If, on the other hand, we have $(x, y, z) \notin \{(\pm 1, 0, 0), (0, \pm 1, 0), (0, 0, \pm 1)\}$ satisfying (2) then we must have $0 \le x^2, y^2, z^2 < 1$, whence $x^4 < x^2, y^4 < y^2, z^4 < z^2$ so that
$x^4 + y^4 + z^4 < x^2 + y^2 + z^2 = 1; \tag{3}$
combining (1) and (3) shows that
$t^4 = (x^4 + y^4 + z^4)^{-1} > 1, \tag{4}$
i.e.
$t(x, y, z) = (x^4 + y^4 + z^4)^{-\frac{1}{4}} > 1 \tag{5}$
is the unique value of $t \in [0, \infty)$ satisfying (1). We now define $\phi:S^2 \to Q^2$ via
$\phi(x, y, z) = t(x, y, z)(x, y, z) = (x^4 + y^4 + z^4)^{-\frac{1}{4}}(x, y, z) \tag{6}$
for $(x, y, z) \in S^2$, i.e. for $x^2 + y^2 + z^2 = 1$. $\phi$ is easily seen to be surjective; indeed, if $(x_1, y_1, z_1) \in Q^2$, setting $s = \sqrt{x_1^2 + y_1^2 + z_1^2}$ we have $s^{-1}(x_1, y_1, z_1) \in S^2$; the corresponding value of $t$ is evidently just $s$ itself, as may be verified by a simple calculation: for $ts^{-1}(x_1, y_1, z_1) \in Q^2$, we have $t^4 s^{-4}(x_1^4 + y_1^4 + z_1^4) = 1$, whence $t^4 s^{-4} = 1$ and so forth. $\phi$ is injective a well: if $(x_1, y_1, z_1), (x_2, y_2, z_2) \in S^2$ and $t_1(x_1, y_1, z_1) = t_2(x_2, y_2, z_2)$ then $t_1^2 = t_1^2(x_1^2 + y_1^2 + z_1^2) = t_2^2(x_2^2 + y_2^2 + z_2^2) = t_2^2$, forcing $t_1 = t_2$ and hence $(x_1, y_1, z_1) = (x_2, y_2, z_2)$ since $t_1, t_2 \ge 0$. This $\phi:S^2 \to Q^2$ is an injective surjection; it is invertible, and the preceding argument shows that, for $(x_1, y_2, z_1) \in Q^2$, $\phi^{-1}(x_1, y_1, z_1) = s^{-1}(x_1, y_1, z_1) \in S^2$; the maps $\phi$ and $\phi^{-1}$ simply project $S^2$ and $Q^2$ onto one another along radial rays emanating from the origin $(0, 0, 0)$. Finally, it is easily seen that both $\phi$ and $\phi_{-1}$ are $C^\infty$, in fact, real analytic maps. Thus $\phi:S^2 \to Q^2$ is a diffeomorphism.
Having established that $Q^2$ and $S^2$ are diffeomorphic, we immediately conclude that the Euler-Poincaré characteristic $\chi(Q^2)$ of $Q^2$ is $2$, since $\chi(S^2) = 2$. We now invoke the Gauss-Bonnet theorem in the form which asserts that for a smooth, compact surface without boundary $\Sigma$ the integral of the gaussian curvature $K$ satisfies
$\int_\Sigma K dA = 2 \pi \chi(\Sigma); \tag{7}$
taking $\Sigma = Q^2$ immediately yields
$\int_{Q^2} K dA = 4 \pi, \tag{8}$
the desired result. QED.
Thanks to the efforts of Messers. Gauss an Bonnet, our real work here lay in establishing, rigorously, that $Q^2$ and $S^2$ are diffeomorphic. Of course, one could in principle attempt to compute $K$ locally directly from the surface definition $x^4 + y^4 + z^4 = 1$, but the calculations become rather messy early on in such a pursuit. I attempted this but soon realized Gauss-Bonnet is a much more tractable approach, easy once one has $\chi(Q^2) = 2$.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
By virtue of Gauss-Bonnet theorem Integra Curvatura ( in Gauss's nomenclature ) is 4 pi in a closed surface. This is the solid angle... as net sum of angle turned in tangent plane on a continuous smooth surface is zero.
