Quick question about a triple integral

[STEMucator]

Homework Statement

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

\iiint\limits_G |xy| dV

Homework Equations

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.

Thanks in advance.

 

[SammyS]

Homework Statement

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

\iiint\limits_G |xy| dV

Homework Equations

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.

Thanks in advance.

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

Your limits of integration are incorrect.

 

[STEMucator]

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

Your limits of integration are incorrect.

? Which ones, also thanks for clarifying that.

 

[SammyS]

? Which ones, also thanks for clarifying that.

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.

 

[STEMucator]

Most of them.

z goes from 0 to 3.

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

etc.

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 ]

:)?

 

[SammyS]

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 ]

:)?

Yes.

Much better !

 

[STEMucator]

Yes.

Much better !

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