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Volume of region R between paraboloid and xy-plane

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data

    So my question is: what is the volume of the region R between the paraboloid 4-x^2-y^2 and the xy-plane?

    2. Relevant equations

    I know how to solve it, it is a triple integral, but how do you find the limits of integration?

    3. The attempt at a solution
    Do I set x=0, or y=0 to find the limits of integration of the triple integral to find the volume ? or? I'm completely lost.
     
  2. jcsd
  3. Nov 11, 2008 #2

    Mark44

    Staff: Mentor

    The paraboloid is defined by an equation, but I don't see one in your problem description.

    A double integral will do the trick. For the limits of integration think about how this paraboloid intersects the x-y plane. The trace in the x-y plane is a circle, right?
     
  4. Nov 11, 2008 #3
    Umm, isn't the equation just z=4-x^2-y^2 ?
    Ok, thanks, I'll try a double integral, so for the limits, I would just set z=0, right? and solve for x and y?
     
  5. Nov 11, 2008 #4
    Ok, for my limits, i got x is between 0 and 2, and y is between 0 and 2...so i did a double integral and got 16/3. Is that right?
     
  6. Nov 12, 2008 #5

    Mark44

    Staff: Mentor

    No.
    If the region over which you are integrating were a square, with 0 <= x <=2 and 0 <= y <= 2, those would be the limits, but it isn't a square. The region R is a circle whose equation is x^2 + y^2 = 4.

    If you divide this region into small squares of area dx * dy, how far up and down (i.e., in the y direction) does a column of squares extend? How far left and right (i.e., in the x direction) do the columns of squares extend? If you can answer these questions correctly, you'll have your limits of integration.
     
  7. Nov 12, 2008 #6
    Would it be plus/minus sqrt(4-x^2) ?
     
  8. Nov 12, 2008 #7

    Mark44

    Staff: Mentor

    Close. A given "vertical" stack ranges from y = -sqrt(4 - x^2) to y = +sqrt(4 - x^2). For the second question, the "vertical" stacks range from x = -2 to x = +2.

    You might notice that I answered with equations, something I'm trying to make you mindful of.

    So now you have your limits of integration. You can make things slightly easier by exploiting the symmetry of your integrand. Because it involves only even powers of x and y, you can reduce your region to just the portion that's in the first quadrant in the x-y plane, introducing a multiplying factor of 4 to correct.

    Things get interesting now because you'll need to integrate some terms that involve square roots and such. These are doable, but tedious. You can save yourself some trouble by converting to polar coordinates, with x = r cos(theta), y = r sin(theta), z = z. The limits of integration are especially simple because the region over which you are integrating is a circular disk. Just remember that dx dy in rectangular coordinates becomes r dr d(theta) in polar coordinates.
     
  9. Nov 12, 2008 #8
    Alright, I used polar coordinates, i think, I got 8pi.
     
  10. Nov 12, 2008 #9

    Mark44

    Staff: Mentor

    So did I.
     
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