Volume between sphere and outside cylinder.

In summary, the conversation discusses the computation of the volume bounded by a sphere and a cylinder, and the difficulties in formulating the integration for this volume. Suggestions are made to use cylindrical or spherical coordinates, or to take slices orthogonal to the x-axis. The conversation also delves into finding the volume of the cylinder inside the sphere using polar coordinates and clarifies the shape and center of the circle in question.
  • #1
peripatein
880
0
Hi,

Homework Statement


I am trying to compute the volume bounded by the sphere x2+y2+z2 = 4 and outside the cylinder x2+y2=2x.


Homework Equations





The Attempt at a Solution


The cylinder's equation is obviously (x-1)2+y2=1, but I am not sure how to formulate the integration for the volume. Using cylindrical coordinates, should the formulation be: ∫(θ=0,2π)dθ∫(r=1,2)dr∫(z=0,sqrt(4-r2))?
 
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  • #2
have you tried to find the volume of the sphere and the cylinder and then subtracting?
making two integrals instead of one, might be easier to see, also use wolframalpha to plot the graph, might give you some hint as to what the ranges are.
also you can use spherical coordinates for one and cylindrical on the other
 
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  • #3
Lucas Mayr said:
have you tried to find the volume of the sphere and the cylinder and then subtracting?
That still leaves the bulk of the difficulty - finding the volume of the cylinder that lies inside the sphere.
Using cylindrical or spherical coordinates here looks messy. How about sticking to Cartesian and taking slices orthogonal to the x axis? Each slice through the cylinder will be a rectangle with segment-of-circle endcaps.
 
  • #4
you could try to separate the two halves, upper and lower,and multiply by 2 at the end, let's say we take the upper half of the cylinder, then,
z goes from 0 to R*Cos∂ (the angule from Z to R),
if you cand find ρ, then 0<θ<2Pi and 0<∂<Pi/2
and you have the volume for the upper part of the cylinder
 
  • #5
Lucas Mayr said:
you could try to separate the two halves, upper and lower,and multiply by 2 at the end, let's say we take the upper half of the cylinder, then,
z goes from 0 to R*Cos∂ (the angule from Z to R),
if you cand find ρ, then 0<θ<2Pi and 0<∂<Pi/2
and you have the volume for the upper part of the cylinder
I think you're overlooking the complexity of the cylinder's shape. What does the line of intersection of cylinder and sphere look like?
 
  • #6
Haruspex, supposing I follow your advice and stick to Cartesian - should I find solid of revolution of cylinder around y-axis? If so, wouldn't it be pi*int(0,2) [x^2*dx*dy/dx], which gives pi*int(0,2) [(x^2-x^3)/sqrt(2x-x^2)]?
 
  • #7
peripatein said:
Haruspex, supposing I follow your advice and stick to Cartesian - should I find solid of revolution of cylinder around y-axis? If so, wouldn't it be pi*int(0,2) [x^2*dx*dy/dx], which gives pi*int(0,2) [(x^2-x^3)/sqrt(2x-x^2)]?

I don't think cylindrical coordinates look at all bad. x^2+y^2=2x has a pretty simple form in cylindrical coordinates. What is it?
 
  • #8
r^2=2rcos(theta), hence r=2cos(theta)?
 
  • #9
peripatein said:
r^2=2rcos(theta), hence r=2cos(theta)?

Right. So that's a circle. What range of theta do you need to cover the circle? Think about graphing it.
 
  • #10
0 to 2pi?
 
  • #11
peripatein said:
0 to 2pi?

Graph it. It's really best if you keep r >=0 everywhere. Figure out which points on the circle correspond to which values of theta.
 
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  • #12
Is it not a circle whose center is at (0,0) and whose radius is 2?
 
  • #13
peripatein said:
Is it not a circle whose center is at (0,0) and whose radius is 2?

No. It's center is at (1,0) and radius 1. Look back at what you got in the problem statement. Which values of theta correspond to which points on the circle?
 
  • #14
Theta = 0 corresponds to (2,0), theta = pi/2 corresponds to (1,1) etc.? Is that what you meant?
 
  • #15
peripatein said:
Theta = 0 corresponds to (2,0), theta = pi/2 corresponds to (1,1) etc.? Is that what you meant?

That's exactly what I mean, but pi/4 corresponds to (1,1), not pi/2. What point corresponds to pi/2?
 
  • #16
Could you please first explain why pi/4 corresponds to (1,1)?
 
  • #17
peripatein said:
Could you please first explain why pi/4 corresponds to (1,1)?

r=2*cos(pi/4)=2*sqrt(2)/2=sqrt(2). So r=sqrt(2), theta=pi/4. That's the point (1,1).
 
  • #18
Okay, I understand why. But I still don't understand how r=2cos(theta) corresponds to a circle whose center is at (1,0) and radius is 1. Would you care to explain, please?
 
  • #19
peripatein said:
Okay, I understand why. But I still don't understand how r=2cos(theta) corresponds to a circle whose center is at (1,0) and radius is 1. Would you care to explain, please?

You wrote that the equation is (x-1)^2+y^2=1 in cartesian coordinates. Doesn't that tell you what it is? The point to figuring out which points on the circle correspond to which values of theta is to figure out a range of theta that will cover the circle once to use as a range in the integration.
 
  • #20
I understand all that, but my question was rather slightly different (I think). Supposing I didn't have that Cartesian equation, and only had r=2cos(theta), how could I have figured out where the center of the circle was, and its radius? Here's a guess: should I have simply needed to find the Cartesian equation first, before attempting to properly describe the circle? I mean, could I have derived C(1,0) and r=1 without having any knowledge of the Cartesian formulation?
 
  • #21
peripatein said:
I understand all that, but my question was rather slightly different (I think). Supposing I didn't have that Cartesian equation, and only had r=2cos(theta), how could I have figured out where the center of the circle was, and its radius? Here's a guess: should I have simply needed to find the Cartesian equation first, before attempting to properly describe the circle? I mean, could I have derived C(1,0) and r=1 without having any knowledge of the Cartesian formulation?

Changing it back to cartesian is the way to go. Recognizing a circle in cartesian coordinates is easy. It's not easy (except for special cases) in polar coordinates.
 
  • #22
In Polar coordinates won't it then be a circle whose center is at the origin? I am not sure how to graphically "translate" my Cartesian equation.
 
  • #23
peripatein said:
In Polar coordinates won't it then be a circle whose center is at the origin? I am not sure how to graphically "translate" my Cartesian equation.

Plot some more values of theta! You've got (2,0) and (1,1) so far. Do some more values for theta until you are convinced it's not a circle around the origin. It's a circle around (1,0). The shape looks the same in either coordinate system.
 
  • #24
Will it also form a circle around (1,0) in Polar coordinates?
 
  • #25
In any case, I believe (0,0) corresponds to theta=pi/2, (+-1/2,+-sqrt(3)/2) corresponds to theta=pi/3. Is that correct? Isn't the range of theta [0,pi/2] U [1.5*pi,2pi]?
 
  • #26
Dick said:
I don't think cylindrical coordinates look at all bad. x^2+y^2=2x has a pretty simple form in cylindrical coordinates.
Sure, but the ranges of integration get nasty when you get into the part where the sphere cuts through the ends of the cylinder.
 
  • #27
peripatein said:
In any case, I believe (0,0) corresponds to theta=pi/2, (+-1/2,+-sqrt(3)/2) corresponds to theta=pi/3. Is that correct? Isn't the range of theta [0,pi/2] U [1.5*pi,2pi]?

Yeah. That range looks good. And ok at pi/3 if you take the + sign. The - sign point isn't on the circle.
 
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  • #28
haruspex said:
Sure, but the ranges of integration get nasty when you get into the part where the sphere cuts through the ends of the cylinder.

Seems to work out ok for me. The domain of the intersection of the cylinder with the xy plane is inside that of the sphere.
 
  • #29
Dick said:
Seems to work out ok for me. The domain of the intersection of the cylinder with the xy plane is inside that of the sphere.
No, I mean where the sphere cuts through the cylinder, defining its ends.
 
  • #30
haruspex said:
No, I mean where the sphere cuts through the cylinder, defining its ends.

As peripetain had (sort of) in the first post, the z values of the ends are where z=+/-sqrt(4-r^2). That's good enough, isn't it?
 
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  • #31
In cylindrical, you have z, r, theta. What will be your integration order?
As a check on the answer, numerically I get about 9.66 for the volume removed by the cylinder, leaving 23.85 in the sphere. Sound right? Still working on the analytic result using Cartesian.
 
  • #32
haruspex said:
In cylindrical, you have z, r, theta. What will be your integration order?
As a check on the answer, numerically I get about 9.66 for the volume removed by the cylinder, leaving 23.85 in the sphere. Sound right? Still working on the analytic result using Cartesian.

I get 9.644049708034451 for the volume removed. So that sounds about right. I integrated dz first, dr second and dtheta third. peripatein had the right general idea in the first post, except that the r factor in the measure was left out, the z limits are wrong, the r limits are wrong and the theta limits are also wrong. It just needs to be fixed. But seriously, if you are still working on the result in Cartesian, I'd give it up unless it's become an obsession. It's HARD to do that way. It's not THAT much of a challenge in cylindrical.
 
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  • #33
Dick said:
I get 9.644049708034451 for the volume removed. So that sounds about right. I integrated dz first, dr second and dtheta third. peripatein had the right general idea in the first post, except that the r factor in the measure was left out, the z limits are wrong, the r limits are wrong and the theta limits are also wrong. It just needs to be fixed. But seriously, if you are still working on the result in Cartesian, I'd give it up unless it's become an obsession. It's HARD to do that way. It's not THAT much of a challenge in cylindrical.
I came to the same conclusion :biggrin:. In the end, I did it with a single integration. In the z plane, the area to be removed is the intersection of two circles. That turns into the sum of two sectors minus the sum of two triangles, so I could write those straight down. Final result is the removal of 16π/3-64/9, leaving 16π/3+64/9. Neatest way to express it is that it leaves 16/9 in the hemisphere it lies in.
 

1. What is the formula for finding the volume between a sphere and an outside cylinder?

The formula for finding the volume between a sphere and an outside cylinder is V = (4/3)πr3 - πr2h, where r is the radius of the sphere and h is the height of the cylinder.

2. How do you determine the radius and height of the cylinder in this calculation?

The radius of the cylinder is equal to the radius of the sphere, and the height of the cylinder is equal to the diameter of the sphere.

3. Can this formula be used for any size sphere and cylinder?

Yes, this formula can be used for any size sphere and cylinder as long as the radius and height are known.

4. What units should be used for the radius and height in this calculation?

The units used for the radius and height should be consistent. For example, if the radius is measured in meters, then the height should also be measured in meters.

5. Can this formula be applied to calculate the volume between any two curved shapes?

No, this formula is specific to finding the volume between a sphere and an outside cylinder. Different shapes will have different formulas for calculating volume.

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