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Homework Help: Fubini's theorem help

  1. Jan 11, 2006 #1
    Hi all,

    I don't fully understand the usage of Fubini's theorem.

    Let's say I have to compute the area of

    [tex]
    \Omega \subset \mathbb{R}^2\mbox{ , }\Omega\mbox{ is bounded with }y = x^2\mbox{ and }y = x + 2
    [/tex]

    [tex]
    \mbox{area(\Omega)} = \iint_{\Omega} 1\ dx\ dy
    [/tex]

    When I draw the situation, I easily find leftmost and rightmost point of this area, which have x-coordinates -1 and 2, respectively.

    Well, Fubini tells us that the integral above can be evaluated as (in this case) two nested one-dimensional integrals. But I have problems with the lower and upper bounds.

    The outer integral is clear, it will be

    [tex]
    \int_{-1}^{2} ...
    [/tex]

    But I'm not sure with the inner one. I know the bounds can't be 0 and 4, because that would give me the area of the whole rectangle.

    So, right solution is this:

    [tex]
    \iint_{\Omega} 1\ dx\ dy = \int_{-1}^{2}\left( \int_{x^2}^{x+2} 1\ dy\right)\ dx = ... = \frac{9}{2}
    [/tex]

    I don't fully get it...why are the bounds of the inner integral [itex]x^2[/itex] and [itex]x+2[/itex]? I know those are the bounding curves, but in Fubini's theorem there is just

    [tex]
    \int_{a_1}^{b_1} \left( \int_{a_2}^{b_2} ...
    [/tex]

    Which I can't connect with the bounds in my example.

    And other question - why is it

    [tex]
    \int_{x^2}^{x+2}
    [/tex]

    and not

    [tex]
    \int_{x+2}^{x^2}\mbox{ ?}
    [/tex]

    Is it because on the interval [itex](-1, 2)[/itex] the function [itex]x+2[/itex] is greater than [itex]x^2[/itex]?

    Thank you very much for clarifying this...
     
    Last edited: Jan 11, 2006
  2. jcsd
  3. Jan 11, 2006 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    You integrate from x^2 to x+2, because x^2 is the lower bound and x+2 the upper bound, as you can easily see from a drawing.

    A way to visualize it is this. What does the inner integral mean? In this case, it is the length of the vertical line between the bounding curves as a function of x. Right? It's simply [itex](x+2)-x^2[/itex] in this case.
    A little more generally, you can use double integrals like [itex]\int_a^b \int_{g(x)}^{h(x)} f(x,y)dydx[/itex] to find the volume of the graph under f(x,y) in which case the inner integral will be the area under the graph obtained by intersecting the graph of f(x,y) with a vertical plane parallel to the yz-plane (or xz-plane, depending on the variable) as a function of x. The inner integral is just a function of x, and it's an area like a standard integral. Then you integrate that along y to get the total volume. If you can visualize that procedure I`m sure you see why the bounds are what they are.
     
    Last edited: Jan 11, 2006
  4. Jan 11, 2006 #3
    Thank you Galileo! I think it's more clear to me now. This was very easy problem, I hope I will understand the three-dimensional as well :)
     
  5. Jan 11, 2006 #4

    benorin

    User Avatar
    Homework Helper

    I'm just curious: does Fubini's Theorem extend to the case of multiple integrals taken w.r.t. different measures?

    Like, say the Lebesgue and counting measures?
     
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