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Calculus Volume Integration

  1. Sep 20, 2009 #1
    I've been trying this equation for few hours now and I can't seem to get the right answer..
    answer is suppose to be 640(pi)/3

    The problem statement, all variables and given/known data

    Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis


    The attempt at a solution
    ``2
    Pi S (9-x^2)^2 - (x^2+1)^2 dx
    ``-2

    ``2
    Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
    ``-2

    Pi[80x - (4x^3)/3] (x=-2 to 2)

    = (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

    that's what I keep getting..
    and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

    thank you in advance
     
  2. jcsd
  3. Sep 20, 2009 #2

    sylas

    User Avatar
    Science Advisor

    Check again your algebra when you expand the squares inside the integration.

    Cheers -- sylas
     
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