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Volume between sphere and outside cylinder.

by peripatein
Tags: cylinder, sphere, volume
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Dick
#19
Dec26-12, 12:46 PM
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Quote Quote by peripatein View Post
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.
peripatein
#20
Dec26-12, 01:00 PM
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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?
Dick
#21
Dec26-12, 01:14 PM
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Quote Quote by peripatein View Post
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.
peripatein
#22
Dec26-12, 01:36 PM
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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.
Dick
#23
Dec26-12, 01:41 PM
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Quote Quote by peripatein View Post
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.
peripatein
#24
Dec26-12, 01:42 PM
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Will it also form a circle around (1,0) in Polar coordinates?
peripatein
#25
Dec26-12, 02:01 PM
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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]?
haruspex
#26
Dec26-12, 03:17 PM
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Quote Quote by Dick View Post
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.
Dick
#27
Dec26-12, 04:19 PM
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Quote Quote by peripatein View Post
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.
Dick
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Dec26-12, 04:22 PM
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Quote Quote by haruspex View Post
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.
haruspex
#29
Dec26-12, 04:32 PM
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Quote Quote by Dick View Post
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.
Dick
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Dec26-12, 05:01 PM
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Quote Quote by haruspex View Post
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?
haruspex
#31
Dec26-12, 09:35 PM
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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.
Dick
#32
Dec26-12, 10:29 PM
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Quote Quote by haruspex View Post
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.
haruspex
#33
Dec27-12, 12:56 AM
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Quote Quote by Dick View Post
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 . 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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