# Describing a Solid in Spherical Coordinates

1. Nov 10, 2008

### daveyman

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
In spherical coordinates, $$x=\rho\sin\phi\cos\theta$$, $$y=\rho\sin\phi\sin\theta$$, and $$z=\rho\cos\phi$$

3. 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!

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Last edited: Nov 10, 2008
2. Nov 10, 2008

### Staff: Mentor

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

3. Nov 10, 2008

### HallsofIvy

Staff Emeritus
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 $\phi= \pi/4$.

4. Nov 10, 2008

### daveyman

Yes, this is correct. Thanks Mark!

5. Nov 10, 2008

### daveyman

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