Describe the region of unit sphere covered by normal Gauss map on the paraboloid $z = x^2+y^2$.
I did this way: Consider the parametrization $X(u,v) = (u,v,u^2+v^2)$. Then $X_u = (1,0,2u)$ and $X_v = (0,1,2v)$.
Then $X_u \times X_v = (-2u,-2v,1)$ and $||X_u \times X_v|| = \sqrt{4(u^2+v^2)+1}$, so $N(u,v) = \frac{(-2u,-2v,1)}{\sqrt{4(u^2+v^2)+1}}$. Since the first and second coordinates can be any real, while the third is always positive, then the image of the paraboloid by Gauss map is the upper open hemisphere.
But the solution I find is done this way:
$S = \{(x,y,z) \in \mathbb{R^3} : f(x,y,z) = x^2+y^2-z = 0\}$, then $N(x,y) = \frac{\nabla f}{||\nabla f||} = \frac{(2x,2y,-1)}{\sqrt{(4(x^2+y^2)+1}}$, then, since third coordinate is always negative, the image of the paraboloid by Gauss map is the lower open hemisphere.
Is one of them wrong? How can the two regions obtained be different? Why can gradient vector be used here?
Thanks.
Note that there is an ambiguity when talking about (normal) Gauss map, in particular when choosing the normal unit vector. Both your choice and the one given by the solution (which is just the negative of what you have) are both acceptable choices.
For example, if you use the chart $$ X(u, v) = (v, u, u^2+v^2),$$ you recover the second answer. If that answer uses $f(x, y,z) = z-x^2-y^2$ they recover what you have.