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Triple Integral in Cartesian Coordinates

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Use a triple integral to find the volume of the solid enclosed by the paraboloid [tex]x=y^2+z^2[/tex] and the plane [tex]x=16[/tex]
    Note: The triple integral must be performed in Cartesian coordinates.

    2. Relevant equations


    3. The attempt at a solution
    I calculated the answer numerically using Mathematica (see attached PDF). I've also included a 3D graph to help with visualization. I calculated the answer to be [tex]128\pi[/tex], but I have no idea how to set up the integral. I'm guessing the argument of the integral will simply be 1, but I don't know how to construct the bounds.
     

    Attached Files:

  2. jcsd
  3. Nov 9, 2008 #2

    gabbagabbahey

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    The parabaloid opens to the right, so x runs from 0 to 16....Try expressing the z bounds in terms of y and x...for a given x and y, what is the minimum z and maximum z-value in the region?
     
  4. Nov 9, 2008 #3
    [tex]z=\sqrt{x-y^2}[/tex]
    The maximum value for z would be [tex]z=\sqrt{16-y^2}[/tex].
    The minimum value for z would be 0.
    Okay, so I have my x and z bounds, but what do I use for my y bound?
     
  5. Nov 9, 2008 #4
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    Last edited: Nov 9, 2008
  6. Nov 9, 2008 #5

    gabbagabbahey

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    Don't you mean [tex]z=\pm \sqrt{x-y^2}[/tex]? :wink:....and you don't want the absolute max and mins for z, you want the max/min for each x,y....which is just [tex]z=- \sqrt{x-y^2}[/tex] for the min, and [tex]z=+ \sqrt{x-y^2}[/tex] for the max....this means that when you integrate over z, you end up with a function of x and y....then you want to integrate over y....what is the boundary of the region in just the x-y plane (ie for z=0)?...what does that mean the max/min of y is for a given value of x?
     
  7. Nov 9, 2008 #6
    Good point about the z bounds. So should I use [tex]y=\pm\sqrt{16}[/tex] for my y bounds?
     
  8. Nov 9, 2008 #7

    gabbagabbahey

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    IF x were always 16, then yes, but for any x, wouldn't you want to use [itex]y=\pm \sqrt{x}[/itex] ? And so when you integrate over y, you get a function of x which you then integrate from 0 to 16....your integral is:

    [tex]V=\int_{\mathcal{V}} dV= \int_0^{16} \int_{-\sqrt{x}}^{\sqrt{x}} \int_{-\sqrt{x-y^2}}^{\sqrt{x-y^2}} dxdydz[/tex]

    ....do you follow?
     
  9. Nov 9, 2008 #8
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    Last edited: Nov 9, 2008
  10. Nov 9, 2008 #9
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  11. Nov 9, 2008 #10
    [tex]\int _0^{16}\int _{-\sqrt{x}}^{\sqrt{x}}\int _{-\sqrt{x-y^2}}^{\sqrt{x-y^2}}1dzdydx = 128\pi[/tex]

    It works beautifully! Thanks again for your help!
     
  12. Nov 9, 2008 #11

    gabbagabbahey

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    No problem :smile:
     
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