Simple questions on the parametrization of the surface $\left\{(x;y;z)\in\mathbb{R}\times\mathbb{R}^{+*}\times\mathbb{R}:x^2+z^2=1/y\right\}$

Question

I am studying alone and I would like to have a feedback on my two first answers in "a)" and "b)" and get help on "c)".

Question:

We have the following parametric surface in $\mathbb{R}^3$, $ \vec{\phi}(r ; \theta ) = \begin{pmatrix} r \cdot cos(\theta)\\ 1/r^2 \\ r \cdot sin(\theta) \end{pmatrix}$, with $r >0 , 0 \leq \theta < 2 \pi $ which is a parametrization I ve found of the surface $S= \left \{ (x;y;z) \in \mathbb{R} \times \mathbb{R}^{+*} \times \mathbb{R} : x^2 + z^2 = 1/y \right \}$
a) Find an orthonormal basis of the tangent plane at all point $( r ; \theta ) $
b) From their deduce a normal vector at all point $( r ; \theta ) $
c) Justify that the parametrization of $S$ you ve found is a bijection

Answer:

a) As we proved here $S$ is a sub manifold of order $2$. Then we can say that the tangent plane of $S$ at any point $x_p$ is $T_{x_0} S = Span \left \{\frac{\partial }{\partial r} \vec{\phi} ; \frac{\partial }{\partial \theta} \vec{\phi} \right \} $ with $\frac{\partial }{\partial r} \vec{\phi} = \begin{pmatrix} cos(\theta)\\ -2/r^3 \\ sin(\theta) \end{pmatrix}$ and $\frac{\partial }{\partial \theta} \vec{\phi} = \begin{pmatrix} -r \cdot sin(\theta)\\ 0 \\ r \cdot cos(\theta) \end{pmatrix} $.

b) Naturally we have that a normal vector $ \vec{v}_{x_0}$ at each point $x_0$ of $S$ is simply given by $\frac{\partial }{\partial r} \vec{\phi} \times \frac{\partial }{\partial \theta} \vec{\phi}$

c) It mays be sound stupid. But how can I prove that $ \vec{\phi}(r ; \theta ) $ is a bijection from $ r>0 ; 0 \leq \theta < 2 \pi $ to $ S $? It seems to me so obvious taht I get difficulty of justifiying it.

Thank you for your help.

Answer 1

I think I ve found a solution for "c)" in all the cases I will be happy to have your feedback on what I ve writen.

c)
To prove that $\vec{\phi} (r ; \theta) $ is a bijection from $r > 0 , 0 \leq \theta < 2\pi $ to $S$ I need to demonstrate that $\vec{\phi} (r ; \theta) $ is injective and surjective.

1- Injectivity : $ f : X \rightarrow Y , \forall a \neq b \in X \Rightarrow f(a) \neq f(b)$
Let take two points $ (r_1; \theta_1) \neq (r_2; \theta_2)$.
In the case it is at least $r_1 \neq r_2 \Rightarrow 1/r_1^2 \neq 1/r_2^2$ (the "$Y$ axis component" of our parametrisation) and so obviously $\vec{\phi}(r_1 ; \theta_1) \neq \vec{\phi}(r_2 ; \theta_2)$.
In the case it is at least $ \theta_1 \neq \theta_2 \neq 0 , \pi$ we have that $sin(\theta_1) \neq sin(\theta_2) $ by the definition of the $sin(.)$ function. Now in the special case that $ \theta_1 = 0 , \theta_2 = \pi $ we can eventually have that the Z and Y axis component are equals but concerning the X axis component we have that $r_1 cos(0) =r_1 \neq -r_2 = r_2 cos( \pi )$.

2- Surjectivity: We need to prove that $ \forall (x';y';z') \in S , \exists (r' ; \theta') $ s.t. $ \vec{\phi}(r'; \theta') = (x';y';z')$
$y' = 1/r'^2 \Rightarrow r' = \sqrt{\frac{1}{y'}} >0$. Thus from here we can find $ \theta' = cos^{-1}(x' \sqrt{y'})$.

"1-" + "2-" $\Rightarrow $ bijection Q.E.D.

Is it correct?

Rem: According to the definition of $S$ it is obvious that $y>0$ in order for $y$ to be a possible coordinate of a point of $S$ along the $Y$ axis.