# Find the parameterization of the surface area given by z = x^2 - 2x + y^2?

## A possible parameterization is $\ast r \ast \left(p , q\right) = \left(1 + p \cdot \cos \left(q\right) , p \sin \left(q\right) , {p}^{2} - 1\right)$, but why?

Jul 19, 2018

Looking at your suggested parameterization, rather than actually finding one:

• $\boldsymbol{r} \left(p , q\right) = \left\langle{\underbrace{1 + p \cos \left(q\right)}}_{= x \left(p , q\right)} , \underbrace{p \sin \left(q\right)} {\setminus}_{= y \left(p , q\right)} , \underbrace{{p}^{2} - 1} {\setminus}_{= z \left(p , q\right)}\right\rangle$

Looking at individual terms of $z \left(x , y\right) = {x}^{2} - 2 x + {y}^{2}$ in terms of the parameterization:

• $\left\{\begin{matrix}{x}^{2} = 1 + 2 p \cos q + {p}^{2} {\cos}^{2} q & q \quad \mathbb{A} \\ - 2 x = - 2 - 2 p \cos q & q \quad \mathbb{B} \\ {y}^{2} = {p}^{2} {\sin}^{2} q & q \quad \mathbb{C}\end{matrix}\right.$

$\mathbb{A} + \mathbb{B} + \mathbb{C} = 1 + \cancel{2 p \cos q} + {p}^{2} {\cos}^{2} q - 2 - \cancel{2 p \cos q} + {p}^{2} {\sin}^{2} q$

$= {p}^{2} \left({\cos}^{2} q + {\sin}^{2} q\right) - 1$

$= {p}^{2} - 1 \textcolor{b l u e}{= z \left(p , q\right)}$

So that seems to work