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Homework Help: Finding volume bounded by paraboloid and cylinder

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the volume bounded by the paraboloid z= 2x2+y2 and the cylinder z=4-y2. Diagram is included that shows the shapes overlaying one another, with coordinates at intersections. (Will be given if necessary)


    2. Relevant equations
    double integral? function1-function2?


    3. The attempt at a solution
    I saw from previous threads involving volumes, but still am lost when I try to do my own problem :\ Most paraboloid involving problems start by changing to polar coordinates- should I do it for this one? I know that at the end it will end up being a double integral, but I am not sure how to set it up.

    physics forums have been a big help. Thanks!
     
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  3. Jul 1, 2011 #2

    vela

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    Are your equations correct because z=4-y2 isn't the equation of a cylinder?
     
  4. Jul 1, 2011 #3

    HallsofIvy

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    Yes, it is. z= 4- y2 is a parabola in the yz-plane and, extended infinitely in the x-direction, is a parabolic cylinder, though not, of course, a circular cylinder.
     
  5. Jul 1, 2011 #4

    vela

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    D'oh!
     
  6. Jul 4, 2011 #5
    thanks for the reply- yes, just like what hallsofivy mentioned, the equations are correct.
    At this point, I am still lost, however. Any other suggestions? Thanks!
     
  7. Jul 5, 2011 #6

    vela

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    As you mentioned in your original post, you want to calculate something like
    [tex]V = \iint\limits_A [z_1(x,y)-z_2(x,y)]\,dy\,dx[/tex]
     
  8. Jul 6, 2011 #7
    Last edited by a moderator: May 5, 2017
  9. Jul 7, 2011 #8

    vela

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    Why are the lower limits 0 for both x and y?
     
  10. Jul 7, 2011 #9
    Well I assumed so, since the problem shape starts from the 0 position for all 3 coordinate systems. Is this approach not correct?
     
  11. Jul 7, 2011 #10

    vela

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    Does the original problem statement say the solid is bounded by the x=0, y=0, and z=0 planes or something equivalent? If it does, your limits look fine. I know the picture suggests this, but you never mentioned it in the original post, nor does it appear in your scan.
     
  12. Jan 15, 2012 #11
    Hey guys, I know this is bringing up an old topic, but I wanted to inquire about something, as well as make sure I approached the final equation correctly.

    To clear up the confusion from vela- yes, I am planning to go with the description saying that since they said to find the volume as indicated in the picture, my limits I set up was going to be from 0.

    Therefore, I went ahead and said

    ∫∫(2x^2-y^2)-(4-y^2),y,0,√2-x^2),x,0,√2).
    After evaluating this, I obtained -pi as my answer. the number makes sense, but the sign is wrong- negative volume is obviously impossible. When I reversed the two functions, I indeed get pi as the answer.

    However, that doesn't make sense- wouldn't the z1 function have to be the function of the paraboloid, and the volume is a subtraction of the cylinder function z2 from z1?

    Any clarification and a check to the final answer will be appreciated. Thanks guys!!
     
  13. Jan 16, 2012 #12

    vela

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    Why do you think that would be the case?
     
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