Let $S = \{(x,y,z) \in \mathbf{R}^3 \mid 4-4x^2-2y^2, x^2+y^2 < 1 \}$. Show that $S$ is a smooth surface.
I'm going to propose two methods that I was taught on how to show that a surface is smooth.
$(1)$ A surface is smooth if all the partials are $C^1$ and the normal vector is non zero at each point i.e there exists a tangent plane at each point on the surface for which the normal is non zero.
Parameterizing $S$ as $\varphi(x,y) =(x,y,4-4x^2-2y^2)$ I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ -8x && -4y \end{bmatrix}$$
from which it can be seen that the partials exist and are continuous.
Now I was thinking that it would satisfy to check whether the Jacobian determinant is non-singular to satisfy the condition for the normals, but the Jacobian determinant cannot be computed for non-square matrices so I guess I have to compute the cross product of the partials.
Computing $\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y} = (8x,4y,1) \ne0$. Which would imply that $S$ is smooth.
$(2)$ The surface is a smooth $2$-dimensional surface if rank$(D\varphi(x,y)) = 2.$
This can be clearly seen from $D\varphi(x,y)$ for example if I reduce the matrix to echelon form I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ 0 && 0 \end{bmatrix}$$
so the rank is clearly $2$ and $S$ is a smooth $2$-dimensional surface.
What I would like to understand is how do these two statements coincide. So how are they equivalent? And also If someone could give me an idea of why the idea of looking at the Jacobian went wrong that would be appreciated.
$\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y}$ is non zero means that the basis vectors are linearly independent so u get a tangent palne which is $\mathbf{R}^2$.
And when u compute the Jacobian and find it is having rank 2 (where u calculated) it means that the linear transformation : U -> $\mathbf{R}^2$.
So both are same. additional ref. Bar page 94