Find the surface area of given this surface(Gauss-Bonnet thm)

Question

There are no solution in my textbook. This book only give me a answer. :(

I don't know why the answer is $4(2\pi + \sqrt3\pi)$

Q) There are a sphere, $S=\{(x,y,z) | x^2 + y^2+z^2=2 \}$ in $R^3$

Then, What is the surface Area of $T =\{(x,y,z) | x^2 + y^2+z^2=2, x-y+\sqrt2z\le\sqrt6 \}$??

Hint) Use the Gauss bonnet theorem.

In my case, when I trying this question

Since the Gaussian Curvature of $S$, $K=\frac{1}{2}$

And all the sum of the external angle is $0$ when we think about the boundary of the surface,

Plus $\int_{\partial S} k_g ds = \pm \sqrt3$

Therefore by Gauss-bonet thm, So Does it have to be $2(2\pi + \sqrt3\pi)$?

Answer 1

You meant $\int_{\partial S}k_g\,ds = \pi\sqrt3$, of course. I don't see how a negative can be explained, as this quantity is orientation-independent. At any rate, we can use classical Euclidean geometry to get the answer — it's the area of a zone of a sphere, and I get $2\pi(\sqrt2)(\sqrt2-\sqrt 6/2) = 2\pi(2-\sqrt3)$, which is what I get with Gauss-Bonnet.