# Solved exercise using Gauss-Bonnet.

I want to solve the following problem using Gauss-Bonnet

Let $$M=\{(x,y,z) \mid z=cos(x^2+y^2), x^2+y^20\}$$ and caluclate $$\int_{M}KdA$$.

I am having some troubles finding a suitable parametrisation for the curve around the region which Id like to integrate the geodesic curvature over.

My own attempt is calculating the whole disc then dividing it by 4, starting as,

$$\gamma_{r}(\theta)=(rcos(\theta),rsin(\theta),cos(r^2))$$

$$\dot{\gamma_{r}}(\theta)=(-rsin(\theta),rcos(\theta),0)$$

$$\ddot{\gamma_{r}}(\theta)=(-rcos(\theta),-rsin(\theta),0)$$

$$N=(-2r^2cos(\theta)sin(r^2),-2r^2sin(\theta)sin(r^2),-r)\frac{1}{4r^2sin^2(r^2)+r^2}$$

$$<\ddot{\gamma},N\times \dot{\gamma}>=\frac{-r}{4r^2sin^2(r^2)+1}$$

Since this is independat of $$\theta$$ we get $$\int_{M}KdA=\frac{\pi}{2}(1+\frac{r}{4r^2sin^2(r^2)+1})$$

• You might want to have your curve to be really at the boundary of the region, and not over the unit circle, and match the definition of the surface, $$γ_r(θ)=(r\cos(θ),r\sin(θ),\cos(r^2)).$$ – LutzL Aug 14 at 15:54
• @LutzL I might..>< – user1 Aug 14 at 15:57
• Isn't there more to the boundary than this? What about $x=0$, $0\le y\le r$, for example? – Ted Shifrin Aug 14 at 18:03
• @TedShifrin I dont understand, to me this looks like deformed disk. Then we just take $\frac{1}{4}$ of it using symmetry. – user1 6 hours ago