Describing a Solid in Spherical Coordinates

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daveyman
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Homework Statement


A solid lies above the cone [tex]z=\sqrt{x^2+z^2}[/tex] and below the sphere [tex]x^2+y^2+z^2=z[/tex]. Describe the solid in terms of inequalities involving spherical coordinates.

Homework Equations


In spherical coordinates, [tex]x=\rho\sin\phi\cos\theta[/tex], [tex]y=\rho\sin\phi\sin\theta[/tex], and [tex]z=\rho\cos\phi[/tex]

The Attempt at a Solution


I have no idea how to do this problem. My attempts have involved converting the two given equations to spherical coordinates, at which point everything is very messy and I don't know where to go next.

I've attached a couple of 3D graphs to help with visualization.

The answer is supposed to be [tex]0\leq\phi\leq\frac{\pi}{4}[/tex] and [tex]0\leq\rho\leq\cos{\phi}[/tex], but this doesn't make much sense to me.

Any help would be great. Thanks!
 

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My recommendation is that you go back and read the problem again!

You say "the cone [itex]z= \sqrt{x^2+ z^2}[/itex] is a complicated cylinder, not a cone. That may be why "everything is messy".

I suspect it was really [itex]z= \sqrt{x^2+ y^2}[/itex]. In that case, the equation in spherical coordinates is [itex]r cos(\phi)= \sqrt{\rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)}= r sin(\phi)[/itex] which reduces to [itex]cos(\phi)= sin(\phi)[/itex]. It should be obvious that that reduces to [itex]\phi= \pi/4[/itex].
 
Mark44 said:
Shouldn't the equation of your cone be [tex]z = \sqrt{x^2 + y^2}[/tex]?

Yes, this is correct. Thanks Mark!
 
HallsofIvy said:
My recommendation is that you go back and read the problem again!

You say "the cone [itex]z= \sqrt{x^2+ z^2}[/itex] is a complicated cylinder, not a cone. That may be why "everything is messy".

I suspect it was really [itex]z= \sqrt{x^2+ y^2}[/itex]. In that case, the equation in spherical coordinates is [itex]r cos(\phi)= \sqrt{\rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)}= r sin(\phi)[/itex] which reduces to [itex]cos(\phi)= sin(\phi)[/itex]. It should be obvious that that reduces to [itex]\phi= \pi/4[/itex].

I didn't see how that reduced but I do now. Thanks!