Spherical coordinates find volume

In summary, the problem asks for the volume of a solid that lies above the cone z= sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = z. The Attempt at a Solution states that if you want to find rho, you need to look at the sphere. However, because you said "below the sphere x^2 + y^2 + z^2 = z," the sphere is the upper limit of integration, and as a result, the value of rho is the maximum value that rho can take on.
  • #1
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Homework Statement


Use spherical coordinates to find the volume of the solid that lies above the cone z= sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = z.


The Attempt at a Solution



I'm having trouble solving for rho (p). I know it starts from 0, and it reaches to the sphere and to the cone. But I don't know where to go from there.
 
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  • #2
Assuming you meant [tex]\rho_o[/tex], the question becomes one of finding [tex]\phi[/tex].
In this case, you have to look at your cone. Essentially, it says [tex]z = r[/tex], where r is in cylindrical coordinates. Now use the conversions to spherical, [tex]z=\rho cos(\phi)[/tex] and [tex] r = \rho sin (\phi)[/tex]. Equating those gives you [tex]cos(\phi) = sin(\phi)[/tex], or [tex] \phi = \pi/4[/tex]. So [tex]0<\phi<\pi/4[/tex].
 
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  • #3
It is a sphere, though. If you move the z over to the left side of the equation, complete the square, you get a sphere with center (0,0,1/2). Also, I need to find rho, not phi.
 
  • #4
Sorry.
[tex] x^2 + y^2 +z^2 = \rho^2[/tex]

[tex] z = \rho cos(\phi)[/tex]

[tex]x^2 + y^2 +z^2 = z[/tex] implies [tex]\rho^2 = \rho cos(\phi)[/tex]

or [tex]\rho = cos(\phi)[/tex]

so [tex]0 < \rho < cos(\phi)[/tex]
 
  • #5
Why are we just looking at the sphere though and not the cone boundary?
 
  • #6
Because you said "below the sphere x^2 + y^2 + z^2 = z" which implies the sphere is the upper bound, so its value for [tex]\rho[/tex] is the maximum value that [tex]\rho[/tex] can take on, thus establishing the upper limit of integration. Hope that made sense :)
 
  • #7
That makes sense...but we are also looking at the cone too. And the value of p for the cone doesn't depend on theta...it is constant. How do we know if we need to divide the triple integral expression into 2 separate ones, one for each different p?
 
  • #8
Okay, this took me a while to find, but here's an example of a time when you need to split the expression for different values of [tex]\rho[/tex].

There are several evident differences between your problem and this one. Most importantly, notice that your solid is "above the cone z= sqrt(x^2 + y^2)" and the example I show is enclosed by the planes. There is a uniquely different value of [tex]\rho[/tex] that the function takes on for different values of [tex]\phi[/tex] namely because the solid that is being integrated must satisfy either being inside a cone or inside a sphere.

Essentially, in the example I give, there are uniquely different values of [tex]\rho[/tex] for different values of [tex]\phi[/tex], whereas in your case, no such thing occurs. If you try to find a different [tex]\rho[/tex], you get what I did in an earlier post:
Knissp said:
Essentially, it says [tex]z = r[/tex], where r is in cylindrical coordinates. Now use the conversions to spherical, [tex]z=\rho cos(\phi)[/tex] and [tex] r = \rho sin (\phi)[/tex]. Equating those gives you [tex]cos(\phi) = sin(\phi)[/tex], or [tex] \phi = \pi/4[/tex]. So [tex]0<\phi<\pi/4[/tex].
Notice that the [tex]\rho[/tex] cancels, thus showing that there is no other unique value to make different bounds of integration.



http://img528.imageshack.us/my.php?image=11fm5.jpg
 
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1. How do I convert from Cartesian coordinates to spherical coordinates?

To convert from 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)

2. What is the formula for calculating the volume of a sphere in spherical coordinates?

The formula for calculating the volume of a sphere in spherical coordinates is V = (4/3)πr³, where r is the radius of the sphere.

3. How do I find the volume of a solid bounded by two spheres using spherical coordinates?

To find the volume of a solid bounded by two spheres using spherical coordinates, you can use the following integral:

V = ∫∫∫ f(r, θ, φ) r² sin(φ) dr dθ dφ

Where f(r, θ, φ) represents the function that defines the solid and the limits of integration are determined by the two spheres.

4. Can I use spherical coordinates to find the volume of a cone?

Yes, you can use spherical coordinates to find the volume of a cone. The formula for the volume of a cone in spherical coordinates is V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone.

5. What are the advantages of using spherical coordinates to find volume?

Using spherical coordinates can be advantageous when dealing with objects that have spherical symmetry, such as spheres, cones, and cylinders. It also simplifies certain integrals and can make calculations easier in certain situations. Additionally, spherical coordinates are often used in physics and engineering applications, making it a useful skill to have as a scientist.

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