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Linear Algebra: Solid Enclosed

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Let E be the solid enclosed by the paraboloid z = x2 + y2 and the plane z = 9. Suppose the density of this solid at any point (x,y,z) is given by f(x,y,z) = x2.


    2. Relevant equations
    x2 + y2 = r2 = 9; r = 3
    ∫∫∫E x2



    3. The attempt at a solution
    The limit of z is given z=9, found r = 3, and θ=2∏

    x = rcosθ
    y = rsinθ
    z = z

    ∫∫∫ r2cos2θ r dzdrdθ

    I got 729∏ / 4 as final answer
     
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2

    LCKurtz

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    I don't think that is correct. Show us your limits and your work.
     
  4. Oct 22, 2012 #3
    z = x2 + y2 and plane z = 9

    I set x2 + y2 = r2
    and found that 0<r<3
    and 0<z<9 since it's given
    and 0<θ<2∏

    x=rcosθ in cylindrical coord.

    f(x,y,z) = x2 --> x2 in cylindrical coord. = (rcosθ)2

    I set the function under a triple integral with restriction above

    so I got ∫∫∫ (rcos)2 rdrdzdθ
     
  5. Oct 22, 2012 #4
    ∫∫z r3 cos2θ drdθ

    9 ∫ r4/4 cos2θ dθ
    (9 * 34)/4 ∫cos2θ dθ
    (9 * 34)/4 ∫(cos2θ + 1)/2 dθ = (9 * 34)/4 *1/2 [ (sin2θ)/2 + θ ] 0<θ<2∏

    I got (9 * 34 * 2∏)/8 = 729∏/4
     
  6. Oct 23, 2012 #5

    LCKurtz

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    Those are not the correct limits. If you are integrating dr first, the value of r depends on z. Your limits describe a cylinder of radius 3 and height 9, which is not what you have. The side of your object is a paraboloid. r goes from r=0 to the r on the paraboloid.
     
  7. Oct 23, 2012 #6
    So the limits to r is from 0 to r = √x2 + y2 or r = √z
    and limit of z is from 0 to 9 ? since the paraboloid begins at 0 and enclosed by the plane at z = 9

    ∫∫∫ r2cos2θ rdrdzdθ
     
  8. Oct 23, 2012 #7

    LCKurtz

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    Yes.
     
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