Finding three orders of integration for a triple integral over unusual region

[mistahkurtz]

Homework Statement

2. The attempt at a solution

It's not hard to find two orders of integration.

(1) Integrate first with respect to x_3, then with respect to x_2, and then with respect to x_1, by dividing D into two regions:

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}\} \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}\}

(2) Integrate first with respect to x_3, then with respect to x_1, and then with respect to x_2, by dividing D into three regions

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}\} \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}\} \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}\}

(3) I'm having difficulty finding how to define D so that I can integrate first with respect to x_2 or x_1, then with respect to x_3, and last with respect to x_1 or x_2. The problem is that the limits of x_3 depend on both x_1 and x_2, 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 x_1 and x_2 are non-negative, let's call it D_1.

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}\}

Since it is also true that 0 \leq x_3 \leq \sqrt{5 - x_1} and 0 \leq x_2 \leq 5 -x_1 - x_3^2, that website recommends defining

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\}

but I don't think that's right. If that were correct, (0.5,4.5,0) would be a point in D_1, which it isn't since if x_1=1, then it must be the case that 0 \leq x_2 \leq 0.5. So basically, I'm stumped. :(

 

[ystael]

Have you tried rotating the coordinate system in the x_1 x_2-plane by 45 degrees?

 

[mistahkurtz]

Have you tried rotating the coordinate system in the x_1 x_2-plane by 45 degrees?

I thought about doing that. But wouldn't the integral no longer be "an integral of a function f(x) over the region D," as the problem explicitly states, but rather "an integral of a function f(\Phi(x)) over the region \phi(D)" where \Phi(x) is the linear transformation that rotates the coordinate system in the x_1 x_2 plane?

EDIT: Yeah, I asked my professor (who wrote the problem), and he says the integral must be written in terms of the coordinates x_1, x_2, x_3.

 

[mistahkurtz]

Well, I think I've progressed a bit towards finding an answer. I think I know how to redefine D_1. This

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\}

is wrong because the upper limit of x_2 should be the minimum of 1 - x_1 and 5 - x_1 - x_3^2 not simply 5 - x_1 - x_3^2. Then, since 1 - x_1 \leq 5 - x_1 - x_3^2 if and only if x_3 \leq 2, we have

D_1 = \{x \in R^3 \mid 0 \leq x_1 \leq 1, 2 < x_3 \leq \sqrt{5 - x_1}, 0 \leq x_2 \leq 5 -x_1 - x_3^2\} \cup \{x \in R^3 \mid 0 \leq x_1 \leq 1, 0 \leq x_3 \leq 2, 0 \leq x_2 \leq 1-x_1\}

Is my reasoning correct? Should I continue doing this - splitting up D into smaller regions and redefining them so the limits of x_3 depend only on x_1?