Find the Cartesian equation of the surface \[\rho^{2}\left(\sin ^{2} \phi-2 \cos ^{2} \phi\right)=1 \]
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Solution
Since
\[\begin{aligned}
&x^{2}+y^{2}=\rho^{2} \sin ^{2} \phi \cos ^{2} \theta+\rho^{2} \sin ^{2} \phi \sin ^{2} \theta\\\\&=\rho^{2} \sin ^{2} \phi
\end{aligned}\]
We have
\[\begin{aligned}
&\rho^{2} \sin ^{2} \phi-2 \rho^{2} \cos ^{2} \phi=1 \quad \\\\&\Longleftrightarrow \quad x^{2}+y^{2}-2 z^{2}=1
\end{aligned}\]
The last equation represents a surface known as ‘one-sheet hyperboloid’. The surface is represented in the next figure
Hyperboloid : \(x^{2}+y^{2}-2 z^{2}=1\)
Home -> Solved problems -> Find the Cartesian equation of the surface
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