Volume integral, how do I find the limits for my integral?

[LHS]

Homework Statement

[PLAIN]http://img9.imageshack.us/img9/4537/unledow.png

Homework Equations

The Attempt at a Solution

Hi, does anyone know how to find the integral that needs to be evaluated here? I can't understand how to find it from the region

edit: Oh and this is from here, not a take home test
http://www.maths.ox.ac.uk/courses/course/12489/material
sheet 3

 

[hunt_mat]

First off I would square everything to obtain:

<br /> a^{2}\leqslant yz\leqslant b^{2},\quad a^{2}\leqslant xy\leqslant b^{2},\quad a^{2}\leqslant xz\leqslant b^{2}<br />

Then I would take the first away from the second to obtain:

<br /> 0\leqslant y(x-z)\leqslant 0<br />

Now what does this say?

 

[LHS]

So does this imply that z<x<z, therefore x=z?

 

[LHS]

I tried the substitution of X=yz, Y=xy, Z=xz
I came out with an answer, but it was extremely messy, due to the Jacobian

 

[hunt_mat]

Well it says that as y>0 then x=z, and you have a\leqslant x\leqslant b

 

[LHS]

ok.. so considering limits for dx,dy,dz

z is between a and b
y is between a^2/z and b^2/z
but what could be do for x?

 

[Ray Vickson]

First off I would square everything to obtain:

<br /> a^{2}\leqslant yz\leqslant b^{2},\quad a^{2}\leqslant xy\leqslant b^{2},\quad a^{2}\leqslant xz\leqslant b^{2}<br />

Then I would take the first away from the second to obtain:

<br /> 0\leqslant y(x-z)\leqslant 0<br />

Now what does this say?

I don't think that is correct. The inequalities a^2 <= xy and a^2 <= xz cannot be
usefully subtracted to get 0 <= x(y-z); for example, 1 <= A and 1 <= B says nothing
about the sign of (A-B). Instead, we can write a^2 <= xy and -b^2 <= -xz, so
*adding* these gives a^2 - b^2 <= x(y-z). Similarly we have
b^2 >= xy and -a^2 >= -xz so
b^2 - a^2 >= x(y-z). Altogether we have -(b^2-a^2) <= x(y-z) <= b^2-a^2, which
does not seem very useful.

RGV