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Double Integrals Advice - Please!

  1. Jan 30, 2006 #1
    Hi, I have a question and it asks me to evaluate the volume generated by the area bounded by y = x^3 and y = x^(1/3) and the function z = x^2.y

    I'm just having a few problems with setting up the ranges of my variables x and y. I drew a sketch of the area in the x-y plane but I'm not sure what my ranges should be. My guess is: 0 <= x <= 1 and 0 <= y <= [x^(1/3) - x^3].

    Help appreciated.
  2. jcsd
  3. Jan 30, 2006 #2


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    As far as the xy-plane is concerned, yes, you can take 0< x< 1 and
    x3< y< x1/3.

    But I am concerned about the lower boundary. Apparently z= x2y is an upper boundary. What is the lower boundary?
    Last edited: Jan 30, 2006
  4. Jan 30, 2006 #3

    Sorry, there seems to be confusion here, on my part at least.

    1.) Your variable ranges are different to mine. Why is mine wrong, and why is yours right?

    2.) What are you talking about, lower boundary? We have an area on the x-y plane and we revolve this to the function z = x^2.y to form a volume. Double Integral (x^2.y) dxdy using our variable ranges will evaluate this volume.
  5. Jan 30, 2006 #4


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    Look at your graph. for each point in the area, y is between the two curves: y is between x3 and x1/3. NOT between 0 and x1/3- x3.

    You said in your first post that you wanted to find the volume of the region generated but didn't say HOW it was generated. I don't understand what you mean by "revolve this to the function z = x^2.y "

    Is this a volume of revolution? If so, what axis are you revolving around?
  6. Jan 30, 2006 #5
    Thanks, I now understand why my range for y was wrong and yours is right. :)

    Sorry, my unclear wording - You have an area in the x-y plane and the volume is generated as: area * height, where height = z(x, y). :. this area is projected in 3-D up to the function z(x, y) to form a volume under z(x, y) with the given x-y plane area as its base.
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