Prove that Gauss map on M is surjective Let $M$ be a closed, orientable, and bounded surface in $\mathbb{R}^3$.
(a) Prove that the Gauss map on $M$ is surjective.
(b) Let $K_+(p) = \max \{0, K(p)\}$.  Show that
$$
\int K_+dA \ge 4\pi.
$$
in the area of $M$.  Do not use the Gauss-Bonnet Theorem to prove this.
 A: For a) there is a simple argument (due to Hilbert IIRC). Pick any unit vector $v$ and imagine a plane with normal vector $v$ with distance $t$ from the origin (imagine $t$ large enough so it is further away from the origin than $M$).
Reduce $t$ until the plane first touches the surface. The point on $M$ where it first touches is a point on $M$ whose image under the Gauss map is $v.$ To make this idea rigourous, you will want to formalize the notion of distance between $M$ and the plane. The compactness of $M$ will ensure the infimum is a minimum.
A: For (b), you need to interpret $\displaystyle\int_M K\,dA$ as the area (counting sign and multiplicity) of the image of the Gauss map. This is the change of variables theorem for integrals, together with the (more or less) definition of $K$ as the determinant of the derivative of the Gauss map.
A: While the answers given are wonderful I thought it would be interesting to look at the first question from the point of view of the Euler characteristic of the surface.
the tangent bundle of the surface is the pull back bundle of the tangent bundle to the sphere via the Gauss mapping. This is easy to see. This means that the pull back of the volume form of the sphere integrates to 2pi times the Euler characteristic of the surface. If the Gauss map were not surjective then the pull back would be an exact form and would integrate to zero. So the Gauss map must be surjective for any surface of non zero Euler charateristic. 
This leaves the question of the torus. Not sure what to do.
A: Thinking about Hilbert's proof it seems that the first touch point of tangency of a given plane to the surface must touch it at a point of non-negative Gauss curvature. For at a point of negative curvature the tangent plane is crossed by the surface - saddle point. Does this give part 2 of the question? On more thought this seems to work. The first touch point of the plane must be tangent at a point of non-negative Gauss curvature because the surface would cut the plane at more than one point at a saddle point. So the Gauss map is surjective onto the sphere just from the points of non-negative curvature. Now pull back the volume form of the sphere and use change of variables. BTW: What can one say about the topology of points of zero Gauss curvature on the surface - say if it has measure zero?
