I'm trying to parametrize the surface given by the equations : $$\frac{x^2}{2}+\frac{y^2}{2}+z^2=1$$ and $x^2+y^2=y$.
I found this function :
$f:[0,1] \times [0,2\pi] \to \mathbb{R}^3$, $$(r,x) \to \left( r \cos(x),r \sin(x),\sqrt{\frac{2-r^2}{2}} \right)$$
but Im not sure at all, can someone verify my answer
On
Hints: Let $S_{1}$ be the surface with equation $\frac{1}{2}(x^{2} + y^{2}) + z^{2} = 1$ and $S_{2}$ the surface with equation $x^{2} + y^{2} = y$.
Since the equation of $S_{2}$ is independent of $z$, the surface $S_{2}$ is some "cylinder" parallel to the $z$-axis. (In fact, $S_{2}$ is a right circular cylinder, as can be checked by completing the square.) You can use polar coordinates ($x = r\cos t$, $y = r\sin t$) to parametrize the "cross section" of $S_{2}$ in the plane $z = 0$ as a polar graph $r = \phi(t)$, so that $x = \phi(t) \cos t$ and $y = \phi(t) \sin t$. Then substitute the resulting functions $(x, y)$ into the equation for $S_{1}$ to express $z$ in terms of $t$.
This works because $S_{2}$ is a cylinder parallel to the $z$-axis and $S_{1}$ is a union of graphs of $z$ as functions of $(x, y)$. A sketch of the surfaces will probably help ensure you've parametrized the entire intersection. It's also a good idea to check that your curves satisfy the equations of $S_{1}$ and $S_{2}$.
Rewrite the equations in cylindrical coordinates
$$x=r\cos\theta,y=r\sin\theta,z.$$
The ellipsoid becomes
$$\frac{r^2}2+z^2=1$$ or$$z=\pm\sqrt{1-\frac{r^2}2}$$ and the cylinder
$$r^2=r\sin\theta,$$or $$r=\sin\theta.$$
Expressing all as a function of $\theta$, the parametric equation of the intersection is
$$\begin{align}x&=\sin\theta\cos\theta,\\ y&=\sin^2\theta,\\ z&=\pm\sqrt{1-\frac{\sin^2\theta}2}. \end{align}$$
Or as a function of $r$, using $\cos\theta=\pm\sqrt{1-\sin^2\theta}$:
$$\begin{align}x&=\pm r\sqrt{1-r^2},\\ y&=r^2,\\ z&=\pm\sqrt{1-\frac{r^2}2}. \end{align}$$
Or as a function of $z$:
$$\begin{align}x&=\pm\sqrt{(2-2z^2)(2z^2-1)},\\ y&=2-2z^2,\\ z&=z. \end{align}$$