Identify the surface whose equation is given in spherical coordinates

**Susaluda** · October 15th 2009, 02:59 PM

Sorry I don't know how to write a nice looking equation on here so, this is the question:

rho^2 (sin^(2) (phi) * sin^(2) (theta) + cos^(2) (phi)) = 9

What surface is this?

**Mush** · October 15th 2009, 03:19 PM

Originally Posted by Susaluda

Sorry I don't know how to write a nice looking equation on here so, this is the question:

rho^2 (sin^(2) (phi) * sin^(2) (theta) + cos^(2) (phi)) = 9

What surface is this?

Just to clarify, this is:

$\rho^2 \big(\sin^2 (\phi) \sin^2(\theta) + \cos^2(\phi)\big) = 9$

**Susaluda** · October 15th 2009, 03:21 PM

Yes, that's right.

**Mush** · October 15th 2009, 03:29 PM

Originally Posted by Susaluda

Yes, that's right.

Are you away that, to convert from spherical to cartesian coordinates, you use the following:

$x = \rho \cos(\phi) \sin(\theta)$

$y = \rho \sin(\phi) \sin(\theta)$

$z = \rho \cos(\theta)$

Some of these bear striking resemblance to your equation, and that might help you to proceed.

**Danny** · October 15th 2009, 03:31 PM

Originally Posted by Mush

Just to clarify, this is:

$\rho^2 \big(\sin^2 (\phi) \sin^2(\theta) + \cos^2(\phi)\big) = 9$

Well, if we consider spherical polar coords (US version)

$<br /> x = \rho \cos \theta \sin \phi,\;\;<br /> y = \rho \sin \theta \sin \phi,\;\;<br /> z = \rho \cos \phi<br />$

then I believe it's a cylinder $y^2 +z^2 = 9$

**Susaluda** · October 15th 2009, 03:33 PM

Wow, yeah, ok I don't know how I didn't realize that :P Thanks to both of you!

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