Finding the volume using spherical coordinates

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Homework Help Overview

The discussion revolves around finding the volume of a solid enclosed by a sphere defined by the equation x² + y² + z² - 2z = 0 and a hemisphere given by x² + y² + z² = 9, with the condition z ≥ 0. The participants are exploring the use of spherical coordinates for this problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss converting the equations to spherical coordinates, identifying the limits for integration, and the challenges in determining the domain for ρ. Some suggest alternative methods, such as using cylindrical coordinates or subtracting volumes of the shapes involved.

Discussion Status

The conversation is ongoing, with participants sharing their attempts and questioning the setup of the problem. Some have provided insights into the geometric interpretation of the shapes involved, while others are still grappling with the integration limits in spherical coordinates.

Contextual Notes

There are indications of confusion regarding the integration limits for ρ, as well as differing opinions on the best coordinate system to use for the problem. The original poster and others are seeking clarity on these points.

Temp0
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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.
 

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