
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 $\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}^{2x}\int_{0}^{1xy}dzdydx$$\displaystyle \rightarrow \int_{0}^{2}\int_{0}^{2x}z\mid_{0}^{1xy}dydx \rightarrow \int_{0}^{2}\int_{0}^{2x}(1xy) dydx$$\displaystyle \rightarrow \int_{0}^{2} y  xy  \frac{y^2} {2}\mid_{0}^{2x}dx \rightarrow \int_{0}^{2} (2x)(2xx^2) (\frac{44x+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.

Hello,
Whew ! I finally spotted the mistake !
You're taking z>0, while, since the plane x+y+z=1 intersects the xyplane, there are some negative z that satisfy the conditions !
Actually, we have z=1xy>12=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 1xy (Wink)
(I haven't done the calculations yet)