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I How to find outer limit of integration for this triple integ

  1. Dec 3, 2016 #1
    Here's a graph and its triple integral. How are the limits of integration for the outer integral [-2,2]? I have no idea how this was found.

    Any help would be appreciated!
     
  2. jcsd
  3. Dec 3, 2016 #2
    What is the boundary at which the cone and spherical cap intersect? As such, between which two x coordinates does the volume lie?
     
  4. Dec 3, 2016 #3
    EDIT: I solved it! I had to substitute z = √(x2 + y2) into x2 + y2 + z2 = 8 then solve for the radius r. Thanks again!

    The boundary has a radius of y = √(4 - x2). I'm not sure what to do with this equation. Hmm..
     
    Last edited: Dec 3, 2016
  5. Dec 3, 2016 #4
    Not quite! The boundary occurs when both surfaces have the same z coordinate, as we can see from the diagram. Equating the z values of the variables in both equations, we see that the x and y coordinates of points on the boundary must satisfy the equation:
    [tex]\sqrt{x^2 + y^2} = \sqrt{8 - x^2 - y^2}[/tex]
    Restricting ourselves to the domain of each square root function, we can simplify this requirement to all points in the domain that satisfy the equation:
    [tex]x^2 + y^2 = 8 - x^2 - y^2[/tex]
    which can be simplified to the equation
    [tex]x^2 + y^2 = 4[/tex]
    This is a circle of radius 2. Do you see how that relates to the boundary for the outer integral?
     
  6. Dec 6, 2016 #5
    Oh, interesting!! Didn't see it that way. Awesome! Thank you for making it even clearer. :)
     
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