# Quick question about a triple integral

1. Nov 26, 2012

### Zondrina

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

Let G be the region bounded by z=x2, z=y2 and z=3. Evaluate :

$\iiint\limits_G |xy| dV$

2. Relevant equations

3. The attempt at a solution

So fixing x and y didn't really give me any useful information. When I fixed z though, I got x=±z and y=±z which forms a square in the xy-plane.

So since my function is continuous and I'm integrating over a square, there shouldn't be any issues whether I pick x or y first.

Now I want to get rid of the absolute value, which seems like more of a logical thing than anything else. So since my region is symmetrical over all my quadrants, could I not just take 4 times the integral in the first quadrant thus eliminating the absolute values? That would cover the entire region if I'm not mistaken thanks to symmetry.

So my iterated integral would be :

$4 * [ \int_{0}^{1} \int_{0}^{z} \int_{0}^{z} xy \space dxdydz ]$

Evaluating that would be easy, I'm just hoping I set it up right. If anyone could validate this I would appreciate it very much.

Last edited: Nov 26, 2012
2. Nov 26, 2012

### SammyS

Staff Emeritus
You certainly can take 4 times the result you get by restricting x & y to the first quadrant.

Your limits of integration are incorrect.

3. Nov 26, 2012

### Zondrina

? Which ones, also thanks for clarifying that.

4. Nov 26, 2012

### SammyS

Staff Emeritus
Most of them.

z goes from 0 to 3.

If z is bounded by $z=x^2\,\$ then x is bounded by $\ x=\sqrt{z}\,,\$ for x in the first quadrant.

etc.

5. Nov 26, 2012

### Zondrina

Ohhh whoops. Thanks man it's getting a bit late, getting sloppy. So it would actually be :

$4 * [ \int_{0}^{3} \int_{0}^{\sqrt{z}} \int_{0}^{\sqrt{z}} xy \space dxdydz ]$

:)?

6. Nov 26, 2012

### SammyS

Staff Emeritus
Yes.

Much better !

7. Nov 26, 2012

### Zondrina

Sweet, thanks for your help Sammy, I got 1 as the answer as I expected ( from the geometry of course ).