Curves on surfaces (differential geometry)

Lee33
Messages
156
Reaction score
0
A few topics we are covering in class are: Gauss map, Gauss curvature, normal curvature, shape operator, principal curvature. I am having difficulty understanding the concepts of curves on surfaces. For example, this problem:

Define the map ##\pi : (\mathbb{R}^3-\{(0,0,0)\})\to S^2## by ##\pi(p)=\frac{p}{||p||}.## Show that if ##\Sigma_R## is the sphere of radius ##R>0##, then the Gauss map of ##\Sigma_R## is ##\pi|_{\Sigma_R}## (which means the map ##\pi## restricted to the surface ##\Sigma_R##.) Compute the shape operator and the Gauss curvature of the sphere.

I don't even know where to start?
 
Physics news on Phys.org
It helps if you write down definitions, what is the Gauss map in question? Can you compute it?
 
I know the Gauss maps a surface in ##\mathbb{R}^3## to the sphere ##S^2,## so ##\pi(p)## is a unit vector for all ##p\in \sum## such that ##\pi(p)## is orthogonal to the surface ##\mathbb{R}^3## at ##p##. Also, we defined the Gauss curvature as: ## K(p) = \kappa_1 \kappa_2 .##
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
5
Views
2K
Replies
9
Views
4K
Replies
13
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Back
Top