Finding the volume using spherical coordinates

In summary: You can't just integrate between those points because the volume of the solid changes when you go outside of the sphere. You need to set up a boundary condition and solve for the volume.
  • #1
Temp0
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0

Homework Statement



Let V be the volume of the solid enclosed by the sphere x^2 + y^2 + z^2 - 2z = 0 ,
and the hemisphere x^2 + y^2 + z^2 = 9 , z ≥ 0. Find V

Homework Equations


Using spherical coordinates:
x^2 + y^2 + z^2 = ρ^2
z = ρcos(ø)

The Attempt at a Solution


So I changed both of them to spherical coordinates, which I get ρ = 3 and ρ = 2 cos (ø). I then attempt to use triple integration, solving ∫∫∫dV where dV = ρ^2 sin(ø) dρdødθ
I find that the domain for θ is [0 , 2∏], and the domain for ø is [0, ∏/2]. However, I'm having trouble finding the domain for ρ. From where to where do I integrate? Any help would be appreciated, thank you in advance.
 
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  • #2
Temp0 said:

Homework Statement



Let V be the volume of the solid enclosed by the sphere x^2 + y^2 + z^2 - 2z = 0 ,
and the hemisphere x^2 + y^2 + z^2 = 9 , z ≥ 0. Find V


Homework Equations


Using spherical coordinates:
x^2 + y^2 + z^2 = ρ^2
z = ρcos(ø)

The Attempt at a Solution


So I changed both of them to spherical coordinates, which I get ρ = 3 and ρ = 2 cos (ø). I then attempt to use triple integration, solving ∫∫∫dV where dV = ρ^2 sin(ø) dρdødθ
I find that the domain for θ is [0 , 2∏], and the domain for ø is [0, ∏/2]. However, I'm having trouble finding the domain for ρ. From where to where do I integrate? Any help would be appreciated, thank you in advance.

There are two better ways to do this than using spherical polars:
(1) Work out the radius of the sphere [itex]x^2 + y^2 + z^2 - 2z = 0[/itex] and subtract its volume from that of the hemisphere.
(2) Use cylindrical polar coordinates.
 
  • #3
The first thing I would do is add 1 to both sides of [itex]x^2+ y^2+ z^2- 2z= 0[/itex] to get [itex]x^2+ y^2+ z^2- 2z+ 1= x^2+ y^2+ (z- 1)^2= 1[/itex] which says that the sphere has center at (0, 0, 1) and radius 1. That is completely contained in the upper hemisphere of a sphere centered at (0, 0, 0) and radius 3. That is why pasmith says you can just find the volume of the hemisphere and subtract from it the volume of the sphere.
 
  • #4
I did it with the method where you subtract the larger volume from the smaller one to find the volume in between, but I want to check with the cylindrical method. I'm having some trouble with it because if I isolate z, then it becomes a huge mess of things and seems impossible to solve.
 
  • #5
Temp0 said:
I did it with the method where you subtract the larger volume from the smaller one to find the volume in between, but I want to check with the cylindrical method. I'm having some trouble with it because if I isolate z, then it becomes a huge mess of things and seems impossible to solve.

You don't want cylindrical coordinates for this problem if you are going to do the integrals. You want to write the smaller sphere ##x^2+y^2+z^2-2z =0## directly in spherical coordinates$$
\rho^2 - 2\rho\cos\phi = 0$$so you have ##\rho=2\cos\phi## for the inner surface and ##\rho=3## for the outer surface. Set it up in spherical coordinates.

[Edit] I see now that the ##\rho## limits answer your original question.
 

What is the formula for finding the volume using spherical coordinates?

The formula for finding the volume using spherical coordinates is V = ∫∫∫ ρ² sin(φ) dρ dφ dθ, where ρ is the radial distance, φ is the polar angle, and θ is the azimuthal angle.

How do spherical coordinates differ from Cartesian coordinates?

Spherical coordinates use a radial distance, a polar angle, and an azimuthal angle to locate a point in 3D space, while Cartesian coordinates use x, y, and z coordinates.

Can spherical coordinates be used for any shape or object?

Yes, spherical coordinates can be used to find the volume of any shape or object, as long as it is a 3D object.

Are there any limitations to using spherical coordinates to find volume?

One limitation of using spherical coordinates is that it may be more difficult to visualize and interpret the results, as compared to using Cartesian coordinates. Additionally, it may be more challenging to apply spherical coordinates to certain irregular or complex shapes.

What are some practical applications of using spherical coordinates to find volume?

Spherical coordinates are commonly used in physics and engineering for calculating the volume of objects such as spheres, cones, and cylinders. They are also used in geographic information systems to map and measure the Earth's surface.

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