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 $$x_1$$?