Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.

Question

Find the equation in spherical coordinates of $x^2 + y^2 – z^2 = 4$.

$$\begin{align} x^2 + y^2 &= r^2\sin^2(\theta)\\ z^2 &= r^2 \cos(\theta) \\ x^2 + y^2 - z^2&=r^2(\sin^2(\theta) - \cos^2(\theta)) = 4 \end{align}$$ Thus, $$-r^2(\cos(2\theta)) = 4$$

Is this right?

Answer 1

The equation given is correct, but we can simplify further. Multiply by $-\sec 2\theta$ and take the square root to solve for $r$ which is $>0$. Thus $r=2\sqrt{-\sec(2\theta)}$.

Note that the result appears to be the square root of a negative number ... unless the secant function is negative. Recalling where the cosine function and thus the secant function is negative, conclude that $\pi/4<\theta<3\pi/4$. This goes along with the hyperboloid shape you might recognize from the rectangular equation.

Answer 2

$x^2 +y^2-z^2=4 \implies \rho ^2-2z^2=4$$

$$ \implies \rho ^2(1-2\cos ^2 ( \phi ))=4$$

$$\implies \rho ^2(\cos 2( \phi ))=4 \implies \rho ^2=4\sec (2\phi)$$