Describing a Solid in Spherical Coordinates

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Homework Help Overview

The problem involves describing a solid that lies above a cone and below a sphere using spherical coordinates. The original poster expresses confusion about converting the given equations into spherical coordinates and mentions that their attempts have resulted in a messy situation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the correct interpretation of the cone's equation, with some suggesting that the original poster may have misidentified it. There is a focus on converting the equations into spherical coordinates and the implications of that conversion.

Discussion Status

Some participants have provided guidance on the potential misinterpretation of the cone's equation, suggesting a clearer path for the original poster to follow. There is an acknowledgment of the confusion surrounding the problem setup, and participants are exploring the implications of the correct equations.

Contextual Notes

The original poster's confusion may stem from the initial problem statement, which includes a potentially incorrect equation for the cone. This has led to a discussion about the proper setup and the resulting inequalities in spherical coordinates.

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

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