Triple Integral

[KLS2111]

Hello,
I am having difficulty setting up the triple integral for the following problem.
Find the volume of the tetrahedron bounded by the planes x+2y+z=2 , x=2y, and x=0, and z=0.

I know I need to find the limits of each in order to set up the integral, but all I have determined is that x is between 0 and 2 y.

I think y goes from - x/2 +1 and x/2. I don't know the upper limit of z or how to set up the integral. Any help you could give would be greatly appreciated. Thanks!

 

[galactus]

The z limits go from z=0 to z=2-x-2y

You have the y limits.

The x limits are from 0 to 1. Because the two lines that make the base of the tetrahedron intersect at x=1.

\(\displaystyle \frac{x}{2}=\frac{-x}{2}+1\)

\(\displaystyle \int_{0}^{1}\int_{\frac{x}{2}}^{\frac{-x}{2}+1}\int_{0}^{2-x-2y}dzdydx=\frac{1}{3}\)

Change the order of integration.

\(\displaystyle \int_{0}^{\frac{1}{2}}\int_{2y}^{2-2y}\int_{0}^{2-x-2y}dzdxdy\)

Here's a diagram of the region I done in Paint. I hope it is OK. I am not an artiste