# Can't understand integral limits in question

1. May 4, 2014

### sa1988

Just a quick one - Have I deciphered the wording of this question properly?

All necessary info is in the embedded image.

To me the limits describe a typical 'pyramid in the corner', but the fact it's an integral over the fucntion 'x' is confusing me. Do I just go with it and generate an answer at the end?

2. May 4, 2014

### BvU

That's going to be difficult, because when integrating over x you don't know the upper bound !
Pyramid in the corner is the picture I get too. I have even sketched it. I notice that when x is known, I know what the integral over the other two represents.

If you want to write the integral properly, it would be a good thing to distinguish between bounds and integration variables. Much clearer. ( In this case not necessary, I realize)

3. May 4, 2014

### sa1988

Well the integral I wrote does come out with a numerical solution. I just don't know if I wrote the correct integral or not.

4. May 4, 2014

### BvU

Ah, I see. You are quite right.

However, if you pick an x, ànd a y, you can easily integrate over z: $\int_{z=0}^{1-x-y} dz$ yielding a simpler expression. Then over y (yielding a triangle area) then finally over x.

So, my nitwit "distinguish" comment is superfluous and I nitwittedly wanted to improve on an integral expression that wasn't wrong in the first place.

5. May 4, 2014

### sa1988

Hmm, I can't see a major world of difference between my $\int_{x=0}^{1-z-y} dx$ and your $\int_{z=0}^{1-x-y} dz$ ???

Ah well, as long as I was correct with my interpretation of the question, and as long as it came to a good answer, I'm happy!

Edit: Oh wait I see it now. If I save the dx integral till last, I'll have far less algebraic havoc to deal with as I go through. Great, all understood, cheers!

6. May 4, 2014

### BvU

You mean $\int_{x=0}^{1-z-y} {\bf x}\, dx$ ?

We agree it's the same thing, but I liked to write it as
$$I = \int_{x=0}^1\, x\, \left ( \int_{y=0}^{1-x} \left ( \int_{z=0}^{1-x-y} dz \right ) dy \right ) dx =\int_{x=0}^1\, x\, \left ( \int_{y=0}^{1-x} 1-x-y \ dy \right ) dx = \int_{x=0}^1\, x\, \left ({1\over 2}(1-x)^2\right )\, dx$$I ended up with 5/24, and you ?

7. May 4, 2014

### sa1988

Ah yeah that's what I meant, sorry I left out the x .

Hmm I get 1/24? I've punched it into mathematica a few different ways and get the same answer with that too.

8. May 4, 2014

### BvU

Roles have reversed! You correctly correct my 5/24. It should be 1/24.
Time to move on to the next exercise...

9. May 4, 2014

### sa1988

Haha, fantastic. Well my main issue was understanding the limits set out in the question, and it turns out everything was ok in the end. Solving it was just a bonus. Thanks!

Oh and I looked back at the method I set out in the original post, and I don't even know why I had the limits set up that way around. Ok it worked but taking dz first should have been obvious! Ah well, living and learning.