# Thread: Triple Integration problem

1. ## Triple Integration problem

I need to integrate x over the region bounded by x=1, y=1, z=1 and x+y+z=2. I am just wondering if the following limits are correct:

1 < z < 2-x-y
1 < y < 3-x
1 < x < 2

I get an answer of -9/8 so I'm assuming that's not right.

2. ## Re: Triple Integration problem

It's been a while but I think the limits are as follows:

$1\leq z\leq 2-x-y$
$1-x\leq y \leq 1$
$0 \leq x \leq 1$

See once we have the limits for z, we can plug z=1 into our equation, solve for y and plot $y=1, x=1, y=1-x$ in the $xy$ plane and see the region clearly to determine the limits for x and y

So we get $y=1-x$ and when we plot it we can see that its the lower limit for y

3. ## Re: Triple Integration problem

I get -1/8 using those limits. Does it make sense for the answer to be negative?

4. ## Re: Triple Integration problem

well with my limits you should have gotten -1/6, but yeah this answer should be positive so something is wrong

5. ## Re: Triple Integration problem

Alright if you reverse my top and bottom limits on z then I think we're good. We both made the mistake of assuming that the region of integration was bounded above by the plane given by x+y+z=2, but that plane is actually the bottom of the region

6. ## Re: Triple Integration problem

Ok. I get 1/8 now. Matlab gives the same answer. Looks right.