# Triple Integral of minimum value

1. Dec 5, 2011

### BioMan789

1. The problem statement, all variables and given/known data
The problem is to take the triple integral over B of min{x,y,z}dV where B = {x,y,z}: o<x<1 , 0<y<1, and 0<z<1. (These are all less than or equal to, I didn't know the notation).

2. Relevant equations

3. The attempt at a solution

What I did was break it up into three cases; where x is the minimum, y is the minimum, and where z is the minimum. I got a value of 1/4 for all three of these, so that leads me to believe the answer for the triple integral is 1/4. Is this correct thinking
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 5, 2011

### SammyS

Staff Emeritus
Hello BioMan789. Welcome to PF.

It would help greatly if you could show just what it was that you integrated. I suspect that you have miisunderstood the meaning of the function, min(x, y, z) .

.

3. Dec 5, 2011

### BioMan789

In between my original post and your response I realized I was integrating incorrectly. I was integrating from 0 to x, 0 to y, 0 to z for each integral instead of 1 to x, 1 to y, 1 to z (see below for clarification). This is what I was integrating.

∫∫∫ x dydzdx (first integral is from x to 1, second is also from x to 1, final is from 0 to 1).
∫∫∫ y dxdzdy (first integral is from y to 1, second is also from y to 1, final is from 0 to 1).
∫∫∫ z dxdydz (first integral is from z to 1, second is also from z to 1, final is from 0 to 1).

4. Dec 5, 2011

### BioMan789

When I worked these integrals out I got 1/12

5. Dec 5, 2011

### SammyS

Staff Emeritus
I'm pretty sure those integrals won't give the correct result.

I've been trying to wrap my head around this for a while. Visualizing this has been a challenge.

$\displaystyle f(x,\,y,\,z)=\min(x,\,y,\,z)=\left\{\matrix{ x\,,\ \ \text{ if }\ x\le y\ \ \text{ and }\ \ x\le z\\ y\,,\ \ \text{ if }\ y\le x\ \ \text{ and }\ \ y\le z\\ z\,,\ \ \text{ if }\ z\le x\ \ \text{ and }\ \ z\le y}\right.$

My suggestion is to write f(x, y, z) as the sum of three functions, fx(x, y, z) + fy(x, y, z) + fz(x, y, z), where
fx(x, y, z) = x when x is the minimum, otherwise it equals zero.

fy(x, y, z) = y when y is the minimum, otherwise it equals zero.

fz(x, y, z) = z when z is the minimum, otherwise it equals zero.​

Then integrate all three functions over the entire cube. But since each is zero over 2/3 of the cube, simply integrate each over the region for which it's not zero. Considering symmetry, each of those integrals will give the same result.
For example, integrate y first for x going from y to 1, then for z going from y to 1, then for y going from 0 to 1.​

I struck part of the line above after looking at the figure in jackmell's post. That line should read:
For example, integrate y first for x going from y to 1, then for z going from 0 to y, then for y going from 0 to 1.​
[/STRIKE]
After further consideration, I withdraw the 1st Edit. and restore the post as it was first written. (This does not imply that jackmell's post was wrong -- only that I misinterpreted his figure.)

Last edited: Dec 5, 2011
6. Dec 5, 2011

### jackmell

I think it's gonna be six integrals over tetrahedrons. Here's a picture of three of them. Got the blue, red and yellow there and then three more behind them I'm not showing. So for example, we have:

$$\mathop\iiint\limits_{\text{Red}} \text{min(x,y,z)}dV=\int_0^1\int_0^x\int_y^x y\;dzdydx$$

Ok, Bioman, you try and do the other five or at least do the yellow one or maybe the blue one would be easier to do first.

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7. Dec 6, 2011

### jackmell

I got those figures by first segregating the cube along diagonal planes shown below. Also, I don't think my wonderful drawings would help when we go to 4D and beyond as in the general problem that was posted a few days ago:

$$\int\cdots\int \text{min}(x_1,\cdots,x_n)dx_1\cdots dx_n$$