Calculus III - Spherical Coordinates

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8. Identify the surface generated by the given equation : \(\rho = - 2\sin \varphi \cos \theta \)

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Let’s first multiply each side of the equation by \(\rho \) to get,

\[{\rho ^2} = - 2\rho \sin \varphi \cos \theta \] Show Step 2

We can now easily convert this to Cartesian coordinates to get,

\[\begin{align*}{x^2} + {y^2} + {z^2} & = - 2x\\ {x^2} + 2x + {y^2} + {z^2} & = 0\end{align*}\]

Let’s complete the square on the \(x\) portion to get,

\[\begin{align*}{x^2} + 2x + 1 + {y^2} + {z^2} & = 0 + 1\\ {\left( {x + 1} \right)^2} + {y^2} + {z^2} &= 1\end{align*}\] Show Step 3

So, it looks like we have a sphere with radius 1 that is centered at \(\left( { - 1,0,0} \right)\).