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Double integral volume of a region bounded by two z planes

  1. Jan 26, 2013 #1
    a) find the volume of the region enclosed by
    z = 1 - y^2 and z = y^2 -1 for x greater or equal to 0 and less than or equal to 2.

    b) would i split up the volume into two integrals, each integral for each z function and then add them together? I also don't know how to find the bounds. Please help with thorough steps
     
  2. jcsd
  3. Jan 26, 2013 #2

    haruspex

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    The simplest is to note that you can split it into two equal volumes (z > 0, z < 0), so you only have to calculate one. But in general, given z > f(x,y) and z < g(x,y), it's as easy to integrate g(x,y)-f(x,y).
    Surfaces like this carve the space into several regions, some of which are infinite. You have to decide which (finite) region is being asked for. Best is to sketch it. In the present case, you have to choose between 1 - y2 < z < y2-1 and 1 - y2 > z > y2-1. Which do you think it is?
     
  4. Jan 26, 2013 #3



    When i sketch the surface it seems to me the limits of integration should be 0 to 2 for x and
    (z+1)^(1/2) to (1-z)^(1/2) for y. would my function inside the integral be
    ( 1 - y^2 - (y^2 - 1) )
     
  5. Jan 26, 2013 #4

    haruspex

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    You need a range for y which does not involve z.
    Pls answer my question: do you think you want the region 1 - y2 < z < y2-1 or 1 - y2 > z > y2-1? Once you've chosen that you'll see what the range for y is.
     
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