# The equation of the surface in cylindrical coordinates: ${r^{2}-2z^{2}=4r\cosθ-8r\sinθ-12z}$ . What is the equation in perpendicular coordinates?

The equation of the surface in cylindrical coordinates:
$${r^{2}-2z^{2}=4r\cosθ-8r\sinθ-12z}$$

i) What is the equation in perpendicular coordinates?
ii) Write it name?

My try:
I put these in the original equation:
$${x=r\cosθ}$$
$${y=r\sinθ}$$
$${z=z}$$
and
$${r=}{\sqrt{x^{2}+y^{2}}}$$
$${θ=\arctan(\frac{y}{x})}$$
$${z=z}$$

So,
$${x^{2}+y^{2}-2z^{2}=4x-8y-12z}$$
$${x^{2}-4x+y^{2}+8y-2z^{2}+12z=0}$$
$${(x-2)^2-4+(y+4)^2-16-2(z-3)^2+18=0}$$
$${\frac{(x-2)^2+(y+4)^2}{2}+(z-3)^2=1}$$

Like a sphere? I'm stuck here
• Center at $(2, -4, 3)$ and $a = \sqrt2, b = \sqrt2, c = 1$ in equation $(x-h)^2 / a^2 + (y-k)^2/b^2 + (z-l)^2 / c^2 = 1$. Jan 31, 2021 at 11:24
You have a mistake in a sign in your last step: $$(x-2)^2+(y+4)^2{\color{red}-}2(z-3)^2=2.$$ So the surface is a hyperboloid of one sheet (see https://en.wikipedia.org/wiki/Quadric): $$\frac{(x-2)^2}{(\sqrt{2})^2}+\frac{(y+4)^2}{(\sqrt{2})^2}-\frac{(z-3)^2}{1^2}=1.$$
• For that it should be $(x-2)^2/2+(y+4)^2/2+(z-3)=0$ without the square in $(z-3)$. Jan 31, 2021 at 12:37