# Thread: The region 0 < x < y < z < 1

1. ## The region 0 < x < y < z < 1

I need to integrate over the region in the title 0 < x < y < z < 1.

Am I correct in understanding that this is the region in x,y,z-space in the first octant with vertex interceptions at (1,0,0), (0,1,0), and (0,0,1) resembling a "pyramid"?

But then the integral of 48*x*y*z over this region must equal one, but I am getting

The integral from x=0 to 1, the integral from y=0 to 1-x, the integral of z=0 to 1 - x - y of 48xyz dz dy dx (this is a triple integral written in "word" form since I don't know latex.

But my answer is 1/15, which is not correct. What am I doing wrong? Thanks!

2. ## Re: The region 0 < x < y < z < 1

Originally Posted by arcketer
The integral from x=0 to 1, the integral from y=0 to 1-x, the integral of z=0 to 1 - x - y of 48xyz dz dy dx (this is a triple integral written in "word" form since I don't know latex. But my answer is 1/15, which is not correct. What am I doing wrong? Thanks!
The region is a pyramid with vertex $(0,0,0),(1,1,1),(0,1,1),(0,0,1)$ . You'll get $\iiint_{R}48xyz\;dxdydz=\int_0^1dx\int_{x}^1dy\int _{y}^148xyz\;dz=\ldots=1$