I am having quite a bit of trouble with triple integrals and I keep coming to this problem.

Find the volume between the plane $\displaystyle x + y + z = 1 $and the $\displaystyle xy $- plane, for $\displaystyle x + y \leq 2, 0 \leq x,$ and $\displaystyle 0 \leq y$

So I started with the bounds which are:

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

$\displaystyle 0 \leq y \leq 2 - x$

$\displaystyle 0 \leq x \leq 2$

And so the triple integral should look like:

$\displaystyle \int_{0}^{2}\int_{0}^{2-x}\int_{0}^{1-x-y}dzdydx$$\displaystyle \rightarrow \int_{0}^{2}\int_{0}^{2-x}z\mid_{0}^{1-x-y}dydx \rightarrow \int_{0}^{2}\int_{0}^{2-x}(1-x-y) dydx$$\displaystyle \rightarrow \int_{0}^{2} y - xy - \frac{y^2} {2}\mid_{0}^{2-x}dx \rightarrow \int_{0}^{2} (2-x)-(2x-x^2)- (\frac{4-4x+x^2} {2}) dx$

After simplifying I ended up with:

$\displaystyle \int_{0}^{2} -x +\frac{x^2} {2} dx \rightarrow -\frac{x^2} {2} + \frac{x^3} {6}$ And After evaluating I ended up with $\displaystyle -\frac{2} {3}$. I am not sure where I went wrong but the answer is supposed to be 1.

I just wanted to say thanks for everyone that has been helping me the past few days with this kind of stuff, I appreciate it a lot.