Volume between sphere and outside cylinder.

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  • #1
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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))?
 

Answers and Replies

  • #2
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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
haruspex
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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
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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
haruspex
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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
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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
Dick
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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
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r^2=2rcos(theta), hence r=2cos(theta)?
 
  • #9
Dick
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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.
 
  • #11
Dick
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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
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Is it not a circle whose center is at (0,0) and whose radius is 2?
 
  • #13
Dick
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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
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Theta = 0 corresponds to (2,0), theta = pi/2 corresponds to (1,1) etc.? Is that what you meant?
 
  • #15
Dick
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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
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Could you please first explain why pi/4 corresponds to (1,1)?
 
  • #17
Dick
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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
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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
Dick
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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
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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
Dick
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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
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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
Dick
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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
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Will it also form a circle around (1,0) in Polar coordinates?
 
  • #25
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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]?
 

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