Describing a Solid in Spherical Coordinates

[daveyman]

Homework Statement

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

Homework Equations

In spherical coordinates, x=\rho\sin\phi\cos\theta, y=\rho\sin\phi\sin\theta, and z=\rho\cos\phi

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 0\leq\phi\leq\frac{\pi}{4} and 0\leq\rho\leq\cos{\phi}, but this doesn't make much sense to me.

Any help would be great. Thanks!

 

[Mark44]

Shouldn't the equation of your cone be z = \sqrt{x^2 + y^2}?

 

[HallsofIvy]

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

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

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

 

[daveyman]

Shouldn't the equation of your cone be z = \sqrt{x^2 + y^2}?

Yes, this is correct. Thanks Mark!

 

[daveyman]

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

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

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

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