Rewriting iterated triple integrals

1. Nov 10, 2007

blitzme

[SOLVED] Rewriting iterated triple integrals

1. The problem statement, all variables and given/known data

Rewrite this integral as an equivalent iterated integral in the five other orders.

2. Relevant equations

$$\int_{0}^{1}\int_{\sqrt{x}}^{1}\int_{0}^{1-y}f(x,y,z) dz dy dx$$

3. The attempt at a solution

Ok, so I have the shape drawn out.
The projection on the XY-axis is the region bounded by half the parabola y=$$\sqrt{x}$$ and the line y=1.

So, $$0 \leq x \leq y^{2}$$, $$0 \leq y \leq 1$$

Or $$0 \leq x \leq 1$$, $$\sqrt{x} \leq y \leq 1$$

The projection on the YZ-axis is the triangle bounded by z=0 and the line z=1-y.

So, $$0 \leq y \leq 1-z$$, $$0 \leq z \leq 1$$

Or $$0 \leq y \leq 1$$, $$0 \leq z \leq 1-y$$

The projection on the XZ-axis is another triangle, bounded by z=0 and the line z=1-x

So $$0 \leq x \leq 1-z$$, $$0 \leq z \leq 1-x$$

Or $$0 \leq x \leq 1$$, $$0 \leq z \leq 1$$

When I use those to make my integrals, I get

1) $$\int_{0}^{1}\int_{0}^{\sqrt{y^{2}}}\int_{0}^{1-x}f(x,y,z) dz dx dy$$

2) $$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-z}f(x,y,z) dy dz dx$$

3) $$\int_{0}^{1}\int_{0}^{1-z}\int_{\sqrt{x}}^{1}f(x,y,z) dy dx dz$$

4) $$\int_{0}^{1}\int_{0}^{1-y}\int_{0}^{1-z}f(x,y,z) dx dz dy$$

5) $$\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{y^{2}}f(x,y,z) dx dy dz$$

I substituted 1 for f(x,y,z) and evaluated the integrals. Number 5 was the only one that was equal to the original (1/12).

What did I do wrong? I would really appreciate your help, thanks.

2. Nov 11, 2007

HallsofIvy

Staff Emeritus
Since the first integral is still with respect to z, you can use exactly the same limits of integration on that integral: 0 to 10- y. Of course, switching x and y, you switch $y= \sqrt{x}$ to $x= y^2$. I assume $\sqrt{y^2}$ was a typo.
$$\int_{y=0}^{1}\int_{x=0}^{y^2}\int_{0}^{1-y}f(x,y,z) dz dx dy$$

Did you notice that the information that $y= \sqrt{x}$ is a boundary has completely disappeared here? Now that can't be right!
You know that z ranges, for fixed y, from 0 up to 1- y. You also know that y can range from 0 up to $\sqrt{x}$. That is, z can range, for fixed x, from 0 up to $1-\sqrt{x}$. Then y can range from 0 up to 1-z.
$$\int_{0}^{1}\int_{0}^{1-\sqrt{x}}\int_{0}^{1-z}f(x,y,z) dy dz dx$$

Just as z did not range from 0 up to 1- x, so x does not range from 0 up to 1- z. x could range from 0 up to $y^2$ and y can range from 0 up to 1-z: x ranges from up to $(1- z)^2$.
$$\int_{0}^{1}\int_{0}^{(1-z)^2}\int_{\sqrt{x}}^{1}f(x,y,z) dy dx dz$$

Same point as in 2 and 3: x ranges from 0 up to $(1-z)^2$
$$\int_{0}^{1}\int_{0}^{1-y}\int_{0}^{(1-z)^2}f(x,y,z) dx dz dy$$

3. Nov 11, 2007

blitzme

First of all, I just want to say thank you, and I appreciate your help.

Yes, $\sqrt{y^2}$ was a typo, and I understand what you did with this one. Also, when I evaluate the integral by plugging in for f(x,y,z) I get the same value as my original answer.

However, for these integrals I get a different answer from my original when I plug in the same f(x,y,z). I understand that I need to have the $$x=y^{2}$$ bound in there in some form, but those integrals don't seem to work out when I check them. That's a valid way of checking the integrals, right? No matter how you write an iterated integral, it should yield the same result for each f(x,y,z).

It wouldn't surprise me if I'm just doing something wrong, but I don't see what.

I think I follow what you did, but are you sure your integrals are right?

Again, thank you for your help!

4. Nov 11, 2007

blitzme

should be
$$\int_{0}^{1}\int_{0}^{1-\sqrt{x}}\int_{\sqrt{x}}^{1-z}f(x,y,z) dy dz dx$$

should be
$$\int_{0}^{1}\int_{0}^{(1-z)^2}\int_{\sqrt{x}}^{1-z}f(x,y,z) dy dx dz$$
and
should be
$$\int_{0}^{1}\int_{0}^{1-y}\int_{0}^{y^2}f(x,y,z) dx dz dy$$

Thanks for helping me out!