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Math Help - use triple integral to find volume of solid bounded by sphere and hyperboloid

  1. #1
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    use triple integral to find volume of solid bounded by sphere and hyperboloid

    bounded below by - x^2 - y^2 + z^2=1 and above by x^2+y^2+z^2=19, for z > 0

    would this be better in cylindrical? z^2 -r^2 =1 and r^2 + z^2 =19

    limits:
    theta: 0 to 2pi
    r: 0 to sqrt 19
    z: sqrt(1+r^2) to sqrt(19-r^2)
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  2. #2
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    I know this is going to sound like crazy talk, especially as it's Saturday night, but this is the sort of problem where the best thing you can do is to work it both ways and see what happens. This is how you develop intuition for math.
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    That does sound crazy, but I will give it a shot Are my limits correct, however?
    Last edited by SNAKE; May 10th 2014 at 09:17 PM.
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    can someone please check my limits to see if they are correct?
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    The sphere and the hyperboloid intersect, but the sphere is above and below the enclosed volume. Do you mean to have a cylindrical wall on the side?

    - Hollywood
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    I think since it says for z > 0, I am looking for the volume inside the cone made by the hyperboloid which is capped by a hemisphere. I don't see where there is a cylindrical wall.
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    Ok, I see it now. I'm not sure what I was thinking before.

    So, yes, cylindrical coordinates would work well. The intersection of the two surfaces is the circle x^2+y^2=9 in the z=\sqrt{10} plane. So your limits are correct except r should go from 0 to \sqrt{10}.

    - Hollywood
    Last edited by hollywood; May 12th 2014 at 07:17 AM. Reason: change z value
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    the region bounded by the paraboloid z=sqrt(x^2+y^2) and above by the sphere x^2+y^2+z^2=6
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    You should start a new thread for a new problem.

    The intersection is the circle x^2+y^2=3 in the z=\sqrt{3} plane.

    So z(x,y) goes from r to \sqrt{6-r^2}. Now we need to define the "floor plan", which is the circle x^2+y^2=3.

    If cylindrical coordinates work for you, your integral is

    \int_0^{2\pi} \int_0^{\sqrt{3}} \int_r^{\sqrt{6-r^2}} r\,dz\,dr\,d\theta

    The extra r is required for cylindrical coordinates.

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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    The region in the first octant bounded above by the cylinder z=1-y^2 and lying between the vertical planes x+y=1 and x+y=3
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    If you go to Calculus - Math Help Forum, there's a button near the bottom that says "+Post New Thread".

    - Hollywood
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    thanks
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    Re: use triple integral to find volume of solid bounded by sphere and hyperboloid

    the region bounded by the paraboloid z=x^2+y^2 and above by the sphere x^2+y^2+z^2=6
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