Smooth surface and parametrization for $\varphi: U → \mathbb{R}^3$ 
Let $U = B(0, 1)$, $\varphi: U → \mathbb{R}^3$, $\varphi(u, v) = (2uv, u^2 − v^2, u^2 + v^2)$ and $S = \varphi(U)$. Determine if $S$ is a smooth surface and if $\varphi$ is the parametrization for it.

So I have that $x=2uv$, $y=u^2-v^2$ and $z = u^2+v^2$. Now $z$ can be expressed as $z=(u-v)^2+2uv =(u-v)^2+x$. Also $y = (u+v)(u-v)$ so $u-v = \frac{y}{u+v}$.
I would get that $z = (\frac{y}{u+v})^2 +x$, thus $\varphi$ would be the correct parametrization?
For the part where they ask if $S$ is smooth I'm not sure what to do. Our course material is pretty awful and I couldn't find similar problems online. I think our course material stated that $S$ would be smooth if it's homeomorphic? So if $\varphi$ is bijective, continuous and has a continuous inverse $\varphi^{-1}$?
 A: Write the equation in vector form: $\vec r(u,v)=(2uv, u^2 − v^2, u^2 + v^2).$ So $\vec r$ maps the disk onto some set of points (that we hope will look like a surface) in $\mathbb R^3.$ By definition, $\vec r$ is $smooth$ if each of the coordinate function partial derivatives is continous, and the normal vector $\vec n=\vec r_u\times \vec r_v$ is non-zero on the interior of $B(0,1).$
The partials are obviously continuous, so all we need to do is check the normal vector:
$\vec r_u(u,v)=2v\vec i+2u\vec j+2u\vec k;\quad \vec r_v(u,v)=2u\vec i-2v\vec j+2v\vec k\Rightarrow$
$\vec n(u,v)=8uv\vec i+4(u^2-v^2)-4(u^2+v^2)\vec k$,
which zero at $(0,0)$ so it is not smooth on $B$. On the other hand, if we avoid this pathological point, and pick another, say $(u_0,v_0)$ and some open ball containing it but not containing $(0,0)$ then $\vec r$ is smooth and we can give an intuitive, geometric reason that $\vec r$ is required to satisfy the above conditions. Suppose that the image of $\vec r $ is indeed a surface. Then, if we note that $(u-u_0)\vec r_u(u_0,v_0)$ and $(v-v_0)\vec r_v(u_0,v_0)$ are vectors tangent to the surface, then (the norm of) their cross-product will be the area of the parallelogram built on these two vectors, and this area will be an approximation of the patch of area on the surface "close" to $(u_0,v_0).$

The idea is then to add up all these areas to get an approximation to the area of the entire surface, and then, by a limiting process (an integral), get the exact area of the surface. But this method fais if the cross-product is zero, because in this case, it may not be possible to define a unique tangent plane (and hence calculate an approximate area) at $(u_0,v_0).$
