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Math Help - integral calculus

  1. #1
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    Question integral calculus

    Find the volume of the region that lies inside z=x^2+y^2 and below the plane z=16.

    Thank you very much.
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  2. #2
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    Quote Originally Posted by kittycat View Post
    Find the volume of the region that lies inside z=x^2+y^2 and below the plane z=16.

    Thank you very much.
    \iiint_V \ dV = \iint_R \int_{x^2+y^2}^{16} dz \ dA = \iint_R x^2+y^2-16 \ dA

    \int_0^{2\pi} \int_0^4 (r^2-14)r\ dr d\theta
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  3. #3
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    Hi,

    Thank you very much for your reply.

    Could you please explain to me why x^2 + y^2 -16 ? Thanks
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  4. #4
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    Quote Originally Posted by kittycat View Post
    Hi,

    Thank you very much for your reply.

    Could you please explain to me why x^2 + y^2 -16 ? Thanks
    It should be,
    16-x^2-y^2
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  5. #5
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    hi Perfecthacker,

    I know

    V=double integral of 16 dA is the volume under z=16
    V= double integral of x^2+y^2 dA is the volume under z= x^2 +y^2

    therefore the desired volume is the difference between them.

    But what I don't understand is
    why the desired volume is the difference between these two double integrals? Could you please explain to me? Thank you very much.
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  6. #6
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    because you want to find the volume that is inside the cone  z = x^2 + y^2 and below the plane  z = 16 .

     \iint_{D} 16 \ dA is the volume under  z = 16 .  \iint_{D} x^2+y^2 \; dA is the volume under  z = x^2 + y^2 .  z = 16 is above that part of the cone. So we want to find the volume that is bounded by the plane and the cone. In other words,  z = 16 is 'higher' than  z = x^2+y^2 . So when you draw in  z = 16 there is still some part of the cone left underneath.

    Last edited by tukeywilliams; July 31st 2007 at 09:37 PM.
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  7. #7
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    Thank you very much turkeywilliam.
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