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Integrals on arbitrary (bounded) domains

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]A = \{(x, y, z) \in \mathbb{R}^n : 0 \lt x \leq 1, 0 \lt y \leq 1 - x^2, 0 \lt z \leq x^2 + y\}[/itex]. Define [itex]f : A \rightarrow \mathbb{R}[/itex] by [itex]f(x, y, z) = y[/itex] for each [itex](x, y, z) \in A[/itex]. Accept that Fubini's theorem is applicable here. Find [itex]\int_A f[/itex].

    2. Relevant equations

    Fubini's theorem must be used. Here, I will give a verbose statement of this theorem. Let [itex]D \subseteq \mathbb{R}^n[/itex] be rectangular (i.e. a Cartesian product of intervals), and suppose [itex]f : D \rightarrow \mathbb{R}[/itex] is integrable. Suppose n = p + q and let [itex]P \subseteq \mathbb{R}^p[/itex], [itex]Q \subseteq \mathbb{R}^q[/itex] such that [itex]D = P \times Q[/itex]. For each [itex]x \in \mathbb{R}^p[/itex], define [itex]f_x : Q \rightarrow \mathbb{R}[/itex] by [itex]f_x(y) = f(x, y)[/itex] and suppose this function is integrable on Q. Now define [itex]F: P \rightarrow \mathbb{R}[/itex] by [itex]F(x) = \int_Q f_x[/itex]. Then F is integrable and [itex]\int_P F = \int_P \int_Q f_x = \int_P \int_Q f(x, y) dy dx = \int_D f[/itex].

    Another very relevant is as follows. Let [itex]A \subseteq \mathbb{R}^n[/itex] be a bounded set, not necessarily rectangular. Let [itex]D \subseteq \mathbb{R}^n[/itex] be rectangular such that [itex]A \subseteq D[/itex]. We are interested in the integral of an integrable function [itex]f : A \rightarrow \mathbb{R}[/itex]. Let [itex]f_0 : D \rightarrow \mathbb{R}[/itex] be defined as [itex]f_0(x) = f(x)[/itex] when [itex]x \in A[/itex], and [itex]f_0(x) = 0[/itex] when [itex]x \in D \setminus A[/itex]. Then [itex]\int_D f_0 = \int_A f[/itex].

    3. The attempt at a solution

    The question asks for an integral of a function defined on a non-rectangular region, so we will begin by enclosing the domain in a rectangle. Let [itex]D = (0, 1] \times (0, 1] \times (0, 2][/itex] so that D is rectangular and [itex]A \subseteq D[/itex]. Extend f to D by defining [itex]f_0 : D \rightarrow \mathbb{R}[/itex] by [itex]f_0(x, y, z) = f(x, y, z) = y[/itex] when [itex](x, y, z) \in A[/itex] and [itex]f_0(x, y, z) = 0[/itex] when [itex](x, y, z) \in D \setminus A[/itex]. By the above result, we get that f0 is integrable on D and [itex]\int_D f_0 = \int_A f[/itex], so now the problem is about an integral on a rectangular region. Thus we can (attempt to) apply Fubini's theorem.

    Let [itex]P, Q, R \subseteq \mathbb{R}[/itex] be the intervals composing D so that [itex]D = P \times Q \times R[/itex]. Define [itex]f_x : Q \times R \rightarrow \mathbb{R}[/itex] by [itex]f_x(y, z) = f_0(x, y, z)[/itex], then define [itex]F : P \rightarrow \mathbb{R}[/itex] by [itex]F(x) = \int_{Q \times R} f_x[/itex]. By Fubini's theorem, we get that

    [tex]\int_P F = \int_{0}^{1} F = \int_{0}^{1} \int_{Q \times R} f_0[/tex]

    Now I'm stuck. I think I've gotten the bounds on the first integral wrong. I think I have the right general idea with the set-up so far but there must be more to it. It has something to do with using 1 - x2 and x2 + y as the bounds on the integral, but I'm really not sure. There are a lot of technicalities and formalities to get out of the way before we can get to the nuts and bolts of the integral.
     
  2. jcsd
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