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Finding the Volume using a triple integral

  1. Feb 15, 2010 #1

    zzz3293

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    1. The problem statement, all variables and given/known data
    Find the volume of the solid bounded by the cylinder x^2+y^2=9 and the planes y+z=5 and z=1


    2. Relevant equations
    None


    3. The attempt at a solution
    My main problem is setting up the integral. So far what I have is 1 as the integrand, my order of integration is dydxdz and my bounds are 0<z<1 and sqrt(9-x^2)<y<5 and I can't figure out x. Is what I have right? What do I need to do? I am really struggling with setting up the bounds for triple integrals
     
    zzz3293, Feb 15, 2010
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  3. Feb 15, 2010 #2

    Mark44

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    Have you drawn a sketch of the solid? Your typical volume element is dy*dx*dz. What are the bounds for y? I.e., y = ?? to y = ??. Then what are the bounds for x?
     
    Mark44, Feb 15, 2010
  4. Feb 15, 2010 #3

    Dustinsfl

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    For the bounds, I obtained 0≤x≤3, -sqrt(9-x^2)≤y≤sqrt(9-x^2), and 1≤z≤5-y. You might want to double check it though.
     
    Dustinsfl, Feb 15, 2010
  5. Feb 15, 2010 #4

    Dustinsfl

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    I forgot to mention I went from 0-3 because I multiplied the integrals by 2.
     
    Dustinsfl, Feb 15, 2010
  6. Feb 15, 2010 #5

    zzz3293

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    I drew it in the xy plane and am looking at it in grapher, I understand the y bounds now, but x and z are still confusing, should x be -3<x<3? I have absolutely no idea how to go about solving for z...
     
    zzz3293, Feb 15, 2010
  7. Feb 15, 2010 #6

    Mark44

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    Yes, assuming that y is as Dustinfl described.
     
    Mark44, Feb 15, 2010
  8. Feb 15, 2010 #7

    Dustinsfl

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    You can use -3 to 3 but since it is a circle you can go from 0 to 3 and multiply by 2 to compensate for -3 to 0. There are 2 planes one is at z=1 and the other is y+z=5. If z=0, then y=5 and if y=0, then z=5, where x can be anything.
     
    Dustinsfl, Feb 15, 2010
  9. Feb 15, 2010 #8

    zzz3293

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    Got it! Now I just need to evaluate which shouldn't be a problem, thanks so much!
     
    zzz3293, Feb 15, 2010
  10. Feb 15, 2010 #9

    Mark44

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    Right. In these types of problems the hardest part is figuring out the limits of integration. Also, Dustinfl's suggestion about exploiting the symmetry is a good one that might reduce the chances of making a calculation error.
     
    Mark44, Feb 15, 2010
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