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Write the equation $x^2+y^2=2y$ in spherical coordinates.

My solution: Using relation between spherical coordinates and rectangular we get:$$(\rho \cos \theta \sin \phi)^2+(\rho \sin \theta \sin \phi)^2=2\rho \sin \theta \sin \phi$$ $$\rho^2 \sin^2 \phi=2\rho \sin \theta \sin \phi$$

However the answer on the book is $\rho \sin \phi=2\sin \theta$. It look like mine but after reduction to $\rho \sin \phi$. Can anyone explain why we can reduce above equation to $\rho \sin \phi$? What about $\rho=0$ and $\phi=0,\pi$?

Would be very grateful for good explanation.

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It is true that $\rho=0$ and $\sin \phi =0$ satisfy the equation, but the equation $\rho \sin \phi = 2 \sin \theta \ (*)$ includes the two conditions anyway.

$\rho = 0$ represents the point $(0,0,0)$, which is obtained in $(*)$ by setting $\rho=0$, $\phi$ as any real number, and $\theta=0$.

$\sin\phi=0$ represents the z axis, which is obtained in $(*)$ by setting $\phi=k\pi$, $\rho$ as any real number, and $\theta=0$.

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  • $\begingroup$ I think in your last sentence something is wrong. If we put in $(*)$ the $\phi=k\pi$ and $\theta=0$ we get $p\sin k\pi=\sin 0$ or $p\cdot 0=0$ ??!! $\endgroup$ – ZFR Oct 15 '17 at 13:21
  • $\begingroup$ sorry I don't see the problem in that; $\rho \cdot 0$ is $0$. $\endgroup$ – Rishi Oct 15 '17 at 13:22

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