Help with parametrization of a Surface and finding tangent plane

Question

So i have Surface defined as:

$$(x^2+y^2+z^2)^3= (x^2−y^2)^2$$ Where $|x|\leq y$

So I was thinking Spherical Coordinates as base, so something like:

$$x=\cos\theta \cos\phi$$ $$y=\cos\theta \sin\phi$$ $$z=sin\theta $$

And since $$|\cos\theta\cos\phi|\leq \ cos\theta\sin\phi$$

Where should i bound $\theta$, $\phi$

Am i using the correct ideas, or am i doing somethign wrong. Any isnight would be helpful. Equation of tangent plane will be kinda trivial, am just having hard tiem with the parametrization. Thank you in advance.

Answer 1

Letting

$$\begin{cases}x&=&r\cos\theta \cos\varphi\\ y&=&r\cos\theta \sin\varphi\\ z&=&r \sin\theta \end{cases}\tag{1}$$

Plugging these relationships into cartesian equation

$$(x^2+y^2+z^2)^3= (x^2−y^2)^2$$

gives, after some computations :

$$r=\pm \cos^2 \theta \cos 2 \varphi \tag{2}$$

(in fact, one can drop the $\pm$ symbol).

Plugging (2) into (1) gives the representation of the surface under the form :

$$\begin{cases}x&=&\cos^3 \theta \cos \varphi\cos 2\varphi\\ y&=&\cos^3\theta \cos 2 \varphi \sin\varphi\\ z&=& \cos^2 \theta \sin\theta \cos 2 \varphi\end{cases}\tag{3}$$

which depends on 2 parameters $\theta$ and $\varphi$.

Answer 2

Your parametrization implies $$x^2+y^2+z^2=1$$

Which alters the equation of surface to $$ (x^2-y^2)^2=1$$

You need to modify your parametrization such that the original surface is not altered.