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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
I request your help

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    $\begingroup$ It is an ellipsoid. $\endgroup$
    – Math Lover
    Jan 31, 2021 at 11:16
  • $\begingroup$ Center at $(2, -4, 3)$ and $a = \sqrt2, b = \sqrt2, c = 1$ in equation $(x-h)^2 / a^2 + (y-k)^2/b^2 + (z-l)^2 / c^2 = 1$. $\endgroup$
    – Math Lover
    Jan 31, 2021 at 11:24

1 Answer 1

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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): $$\frac{(x-2)^2}{(\sqrt{2})^2}+\frac{(y+4)^2}{(\sqrt{2})^2}-\frac{(z-3)^2}{1^2}=1.$$

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  • $\begingroup$ Thank you for reply. But, you said it is "hyperboloid". If the denominator of (z-3) is "1", then isn't it "Elliptic paraboloid"? $\endgroup$
    – gunza
    Jan 31, 2021 at 12:31
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    $\begingroup$ For that it should be $(x-2)^2/2+(y+4)^2/2+(z-3)=0$ without the square in $(z-3)$. $\endgroup$ Jan 31, 2021 at 12:37

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