I have been trying to solve the problem below, but I have kept arriving at the same answer. I think I set up the bounds incorrectly, but I do not have a sure way of knowing:
Find the volume of solid bounded by 3 coordinate planes, bounded above by plane x+y+z=2 and bounded below by z=x+y.
The integration problem is then
$V = \displaystyle \int_{x} \int_{y} \int_{z=x+y}^{2-x-y} 1dzdydx$, which can be simplified to
$V = \displaystyle \int_{x} \int_{y} (2-x-y-(x+y))dydx$, or
$V = \displaystyle \int_{x} \int_{y} (2-2x-2y)dydx$.
I calculated the bounds for x to be $0 \leq x \leq 2$ and for y to be $0 \leq y \leq 2-x$. Thus, my construction of the integral becomes $V = \displaystyle \int_{x=0}^{2} \int_{y=0}^{2-x} (2-2x-2y)dydx$ which evaluates to $-\frac{4}{3}$. This not correct. The integral is actually $\frac{1}{3}$.