(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The region R in 3D is cut from the first octant (x,y,z >= 0) by the plane X+Z = 1, Y+2Z = 2.

Set up the volume in all 6 ways in rectangular coordinates.

Then evalute the volume in two of these ways.

Make sure to specify limits of integration in every case.

2. Relevant equations

I figure Y+2Z = 2 is the upper bound for Z and Z=0 is the lower bound.

Here's a rough sketch I drew of the base on X-Y plane, and the 3-D of what I imagined it to look like:

[PLAIN]http://img576.imageshack.us/img576/2599/calc3.png [Broken] [/PLAIN]

3. The attempt at a solution

So I put the triple integration in:

[tex]

V(R) = \int_0^1{\mathrm \int_(2X)^(2-2X){\mathrm \int_0^(1-(1/2)Y){\mathrm }{\mathrm d}Z}{\mathrm d}Y}{\mathrm d}X

[/tex]

When I work this out, I keep getting a negative value!

I know the answer is suppose to be 2/3 because a rectangular pyramid's volume = 1/3 * height * base area, which the base area is 1x2 times 1 height * 1/3 = 2/3.

But I cant seem to manipulate it anyway to get it. Where am I going wrong? Am I setting the wrong plane as the Z bound?

Thanks for any help.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Triple Integral for volume

**Physics Forums | Science Articles, Homework Help, Discussion**