Volume between sphere and outside cylinder.

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  • #26
haruspex
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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
Dick
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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]?

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
Dick
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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
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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
Dick
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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, isnt it?
 
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  • #31
haruspex
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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
Dick
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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
haruspex
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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.
 

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