
#19
Dec2612, 12:46 PM

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#20
Dec2612, 01:00 PM

P: 812

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
Dec2612, 01:14 PM

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#22
Dec2612, 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.




#23
Dec2612, 01:41 PM

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#24
Dec2612, 01:42 PM

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Will it also form a circle around (1,0) in Polar coordinates?




#25
Dec2612, 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]?




#26
Dec2612, 03:17 PM

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Dec2612, 04:19 PM

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#28
Dec2612, 04:22 PM

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#29
Dec2612, 04:32 PM

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Dec2612, 05:01 PM

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#31
Dec2612, 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. 



#32
Dec2612, 10:29 PM

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#33
Dec2712, 12:56 AM

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