Describing a Solid in Spherical Coordinates

In summary, the cone in question has an equation of z=\sqrt{x^2+y^2} which reduces to z=\sqrt{x^2+sin^2(\phi)}.
  • #1
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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  • #2
Shouldn't the equation of your cone be [tex]z = \sqrt{x^2 + y^2}[/tex]?
 
  • #3
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].
 
  • #4
Mark44 said:
Shouldn't the equation of your cone be [tex]z = \sqrt{x^2 + y^2}[/tex]?

Yes, this is correct. Thanks Mark!
 
  • #5
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!
 

1. What are spherical coordinates and how are they used to describe a solid?

Spherical coordinates are a system of coordinates used to locate points on a sphere or a three-dimensional space. They consist of three values: radius, azimuth, and inclination. They are often used to describe the location and shape of a solid object in three-dimensional space.

2. How do I convert Cartesian coordinates to spherical coordinates?

To convert Cartesian coordinates (x, y, z) to spherical coordinates (r, θ, φ), you can use the following formulas:
r = √(x² + y² + z²)
θ = arctan(y/x)
φ = arccos(z/r)
Note: make sure to use the correct units for each value (e.g. radians or degrees).

3. Can spherical coordinates be used to describe any type of solid?

Yes, spherical coordinates can be used to describe any type of solid, as long as the shape can be approximated by a sphere or a combination of spheres. This includes shapes such as spheres, cylinders, cones, and ellipsoids.

4. How are spherical coordinates used in physics and engineering?

Spherical coordinates are commonly used in physics and engineering to describe the location and orientation of objects in three-dimensional space. They are particularly useful in applications involving rotation and symmetry, such as in celestial mechanics, fluid dynamics, and electromagnetic fields.

5. Are there any limitations to using spherical coordinates to describe a solid?

While spherical coordinates are a useful tool for describing many types of solids, they do have some limitations. For example, they are not well-suited for describing irregular or asymmetric shapes. Additionally, spherical coordinates are not always the most efficient or convenient system for calculations, so other coordinate systems may be preferred in certain situations.

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