Finding the volume using spherical coordinates

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The discussion focuses on finding the volume of a solid enclosed by two spheres using spherical coordinates. The inner sphere is represented by the equation x^2 + y^2 + z^2 - 2z = 0, which translates to ρ = 2cos(ø) in spherical coordinates, while the outer hemisphere has ρ = 3. The integration setup involves triple integration with the limits for θ being [0, 2π] and for ø being [0, π/2]. Participants suggest alternative methods, such as subtracting the volume of the smaller sphere from the larger hemisphere or using cylindrical coordinates, although the latter is deemed less effective for this problem. The discussion concludes with clarification on the limits of integration for ρ, confirming that they are indeed 2cos(ø) to 3.
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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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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 x^2 + y^2 + z^2 - 2z = 0 and subtract its volume from that of the hemisphere.
(2) Use cylindrical polar coordinates.
 
The first thing I would do is add 1 to both sides of x^2+ y^2+ z^2- 2z= 0 to get x^2+ y^2+ z^2- 2z+ 1= x^2+ y^2+ (z- 1)^2= 1 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.
 
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.
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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