# Change the order of triple integration

1. Nov 23, 2009

### coco87

1. The problem statement, all variables and given/known data
Rewrite $$\int_{0}^{2}\int_{0}^{y^3}\int_{0}^{y^2}dzdxdy$$ as an integral with order $$dydzdx$$.

2. Relevant equations
N/A

3. The attempt at a solution
Honestly, I got as far as sketching it:

and after sketching it, I'm lost...

I can't figure out how to set $$z$$ or $$y$$, but I'm fairly sure that $$x$$ is $$[0,8]$$.

could anyone possibly offer some help?

Thanks

2. Nov 23, 2009

### Staff: Mentor

Is this all you're given in the problem? Just change the order of integration? I'm concerned that you might be omitting some information in the problem.

BTW, I've always found it to be easier to include the variable in one of the limits of integration, when I'm dealing with iterated integrals.
$$\int_{y = 0}^{2}\int_{x = 0}^{y^3}\int_{z = 0}^{y^2}dz~dx~dy$$

3. Nov 23, 2009

### coco87

Mark44:

Thank you for your tip :)

I'm actually not leaving anything out at all. Many of these questions are vague (and rather annoying). I think I might have it, but am not sure:

$$\int_{0}^{8}\int_{0}^{\sqrt[3]{x}}\int_{0}^{\sqrt{z}}dydzdx$$. However, this seems way to easy to be true...

4. Nov 23, 2009

### Staff: Mentor

Well, you can check. both integrals represent the volume of the region, so both integrals should produce the same value.

5. Nov 23, 2009

### coco87

Mark44:

Hmm, how would I check without a function to integrate?

6. Nov 23, 2009

### Staff: Mentor

The function is 1.