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Math Help - Iterated Intergal

  1. #1
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    Iterated Intergal

    Find the volume of the solid inside the cylinder x^2+y^2 = 2y and bounded above and below by the sphere x^2 + y^2 + z^2 = 4.

    Thanks for visiting through my problems. I need to get the solution but i do not have any idea on how to solve it. Please help. Thanks
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  2. #2
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    Quote Originally Posted by guess View Post
    Find the volume of the solid inside the cylinder x^2+y^2 = 2y and bounded above and below by the sphere x^2 + y^2 + z^2 = 4.

    Thanks for visiting through my problems. I need to get the solution but i do not have any idea on how to solve it. Please help. Thanks
    By completing the square, that cylinder is x^2+ (y- 1)^2= 1 which has axis (0, 1, z) which is slightly of center from the sphere. I think the first thing I would do is "translate" the origin to (0, 1, 0)- replace y by y'= y- 1 so that y= y'+ 1 and the equation of the cylinder is x^2+ y'^2= 1 and the equation of the sphere is x^2+ (y'+ 1)^2+ z^2= 4 or x^2+ y'^2+ z^2+ 2y= 3.

    Now the cylinder has its axis at (0, 0, z) and you can use cylindrical coordinates: z will run from the part of the sphere at the bottom, z= -\sqrt{3- x^2- y^2- 2y}= -\sqrt{3- r^2- 2rsin(\theta)}, to the part at the top, z= \sqrt{3- x^2- y^2- 2y= \sqrt{3- r^2- 2rsin(\theta)} while r and \theta range from 0 to 1 and 0 to 2\pi. The volume is given by
    \int_{\theta= 0}^{2\pi}\int_{r= 0}^1\int_{z=-\sqrt{3- r^2- 2rsin(\theta)}}^{\sqrt{3- r^2- 2rsin(\theta)}} r  dz dr d\theta
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  3. #3
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    Hi. Here's an alternate way just for fun. Do the complete the square thing, then convert x^2+(y-1)^2=1 to polar coordinates to obtain r=2\sin(t), and the sphere into cylindrical coordinates z^2=\sqrt{4-r^2}, then the integral is:

    4\int_0^{\pi/2}\int_0^{2\sin(t)}\int_0^{\sqrt{4-r^2}} rdzdrdt which is the same answer as HallsofIvy above, about 9.6 which is in the ball park since the volume of the whole cylinder of length 4 is 4\pi\approx 12.6
    Last edited by shawsend; March 30th 2010 at 04:12 AM.
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