Parametrisation of surfaces

Question

For the surface below find a parametrisation, compute $\vec{r}_α, \vec{r}_β, \vec{r}_α × ~\vec{r}_β$ and find tangent plane at $(1,1,0)$

$x^2+y^2−z^2 = 2y+2z$ where $−1 ≤ z ≤ 0$

Answer 1

\begin{align*} x^2+y^2-z^2 &= 2y+2z \\ x^2+(y-1)^2 &= (z+1)^2 \\ (x,y-1,z+1) &= (\beta \cos \alpha,\beta \sin \alpha,\beta) \\ (x,y,z) &= (\beta\cos \alpha,1+\beta\sin \alpha,\beta-1) \\ \mathbf{r}_{\alpha} &= \beta(-\sin \alpha,\cos \alpha,0) \\ \mathbf{r}_{\beta} &= (\cos \alpha,\sin \alpha,1) \\ \mathbf{r}_{\alpha} \times \mathbf{r}_{\beta} &= \beta (\cos \alpha,\sin \alpha,-1) \\ \end{align*}

$$(1,1,0)=(\beta \cos \alpha,1+\beta \sin \alpha,\beta -1) \implies (\alpha,\beta)=(0,1) \implies \mathbf{r}_{\alpha} \times \mathbf{r}_{\beta}=(1,0,-1)$$

The equation of plane is

$$1\times (x-1)+0\times (y-1)-1\times (z-0)=0$$

$$\fbox{$x-z=1$}$$

Answer 2

We have

$$x^2+(y^2-2y)-(z^2+2z)=x^2+(y-1)^2-(z+1)^2=0$$

We let $x=v\cos(u)$, $y=1+v\sin(u)$, and $z=v-1$.

Then, we have

$$\vec r=\hat x v\cos(u)+\hat y (1+v\sin(u))+\hat z (v-1)$$

and

$$\frac{\partial \vec r}{\partial u}=-\hat x v\sin(u)+\hat y v\cos(u)$$

$$\frac{\partial \vec r}{\partial v}=\hat x \cos(u)+\hat y \sin(u)+\hat z$$

Finally, the normal to the surface is

$$\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}=v(\hat x \cos(u)+\hat y \sin(u)-\hat z)$$

Answer 3

Hint:

The surface is

$$x^2+(y-1)^2-(z+1)^2=0\iff z=\pm\sqrt{x^2+(y-1)^2}-1 =$$

a double cone, so you can parametrize each sign in a similar way, say

$$r(u,v):=\left(\,u,\,v,\,\sqrt{u^2+(v-1)^2}-1\,\right)\;,\;\;0\le u,\,v\le1$$