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Funny looking equation for a paraboloid.

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the volume between 2 paraboloids.

    2. Relevant equations

    z = 4x^2+8y^2
    z= 30-x^2-y^2

    3. The attempt at a solution

    So I switched the variables to polar coordinates.
    z = 30-r^2

    Now I want to solve for r. However I get a very ugly looking value for r that does not integrate very well.
  2. jcsd
  3. Apr 17, 2010 #2


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    Why do you want to solve that for r? You want to do a double integral of zupper - zlower over the common xy domain. Figure out the equation of that domain and it might suggest a change of variables to you.
  4. Apr 17, 2010 #3
    I recommend two things:

    1. No change of variables or coordinate systems, it just makes things more intricate.
    2. Find the bounds first. They aren't as complex as you would imagine.

    All of this assuming, you will use a double integral.

    To find the upper paraboloid just evaluate the two at a point inside of the bounds you find.
  5. Apr 17, 2010 #4
    To find the bounds, I would set z=0 and then solve for x and then y. For example...y = sqrt(24-x^2).
    Last edited: Apr 17, 2010
  6. Apr 17, 2010 #5
    Instead of solving for y, if you keep the x's and y's on the same side, you should see a familiar function in there.

    Attached Files:

  7. Apr 17, 2010 #6
    equation of a circle of sorts?
  8. Apr 17, 2010 #7
    yeah, I added an attachment, to my previous post, that shows the two paraboloids from the top. If you find the function for the boundary shape, you can find your bounds of x and y. Remember, you only need xmin xmax ymin and ymax
  9. Apr 18, 2010 #8


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    No. You will have an ellipse shaped boundary which does not give constant limits. And a substitution similar to, but not exactly the same as, polar coordinates will be very helpful. You need to set the two z values equal to get the equation of the ellipse.
  10. Apr 18, 2010 #9
    LCKurtz is right, how could I be so silly, you can still get to the ellipse in cartesian, but for the bounds you will definitely need polar coordinates of some sort.
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