- #1
mistahkurtz
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Homework Statement
2. The attempt at a solution
It's not hard to find two orders of integration.
(1) Integrate first with respect to [tex]x_3[/tex], then with respect to [tex]x_2[/tex], and then with respect to [tex]x_1[/tex], by dividing D into two regions:
[tex]D = \{x \in R^3 \mid -1 \leq x_1 < 0, -\sqrt{1-x_1^2} \leq x_2 \leq \sqrt{1-x_1^2}, -\sqrt{5 - x_1 - x_2} \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex] [tex]\cup \{x \in R^3 \mid 0 \leq x_1 \leq 1, -\sqrt{1-x_1^2} \leq x_2 \leq 1-x_1, -\sqrt{5 - x_1 - x_2} \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex]
(2) Integrate first with respect to [tex]x_3[/tex], then with respect to [tex]x_1[/tex], and then with respect to [tex]x_2[/tex], by dividing D into three regions
[tex]D = \{x \in R^3 \mid -1 \leq x_2 < 1, -\sqrt{1-x_2^2} \leq x_1 < 0, -\sqrt{5 - x_1 - x_2} \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex] [tex]\cup \{x \in R^3 \mid 0 \leq x_2 \leq 1, 0 \leq x_1 \leq 1-x_2, -\sqrt{5 - x_1 - x_2} \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex] [tex]\cup \{x \in R^3 \mid -1 \leq x_2 \leq 0, 0 \leq x_1 \leq \sqrt{1-x_2^2}, -\sqrt{5 - x_1 - x_2} \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex]
(3) I'm having difficulty finding how to define D so that I can integrate first with respect to [tex]x_2[/tex] or [tex]x_1[/tex], then with respect to [tex]x_3[/tex], and last with respect to [tex]x_1[/tex] or [tex]x_2[/tex]. The problem is that the limits of [tex]x_3[/tex] depend on both [tex]x_1[/tex] and [tex]x_2[/tex], and I can't seem to manipulate the inequalities correctly to give me what I want.
I tried following the method used in Example 5 here (http://www.math.umn.edu/~nykamp/m2374/readings/tripintex/) in order to redefine just the subset of D for which [tex]x_1[/tex] and [tex]x_2[/tex] are non-negative, let's call it [tex]D_1[/tex].
[tex]D_1 = \{x \in R^3 \mid 0 \leq x_1 \leq 1, 0 \leq x_2 \leq 1-x_1, 0 \leq x_3 \leq \sqrt{5 - x_1 - x_2}\}[/tex]
Since it is also true that [tex]0 \leq x_3 \leq \sqrt{5 - x_1}[/tex] and [tex]0 \leq x_2 \leq 5 -x_1 - x_3^2[/tex], that website recommends defining
[tex]D_1 = \{x \in R^3 \mid 0 \leq x_1 \leq 1, 0 \leq x_3 \leq \sqrt{5 - x_1}, 0 \leq x_2 \leq 5 -x_1 - x_3^2\}[/tex]
but I don't think that's right. If that were correct, (0.5,4.5,0) would be a point in [tex]D_1[/tex], which it isn't since if [tex]x_1=1[/tex], then it must be the case that [tex]0 \leq x_2 \leq 0.5[/tex]. So basically, I'm stumped. :(
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