1. ## Triple Integrals

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

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

So I started with the bounds which are:
$0 \leq z \leq 1 - x - y$
$0 \leq y \leq 2 - x$
$0 \leq x \leq 2$

And so the triple integral should look like:

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

After simplifying I ended up with:

$\int_{0}^{2} -x +\frac{x^2} {2} dx \rightarrow -\frac{x^2} {2} + \frac{x^3} {6}$ And After evaluating I ended up with $-\frac{2} {3}$. 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.

2. Hello,

Whew ! I finally spotted the mistake !

You're taking z>0, while, since the plane x+y+z=1 intersects the xy-plane, there are some negative z that satisfy the conditions !

Actually, we have z=1-x-y>1-2=-1, because x+y<2 (large inequalities, but I'm too lazy to do the latex)

So try it with the boundaries for z : -1 and 1-x-y

(I haven't done the calculations yet)