Exercise text$\ \ \ $ Calculate $$\ \ \iiint_R y \ dxdydz\ \ $$ where $R$ is the cube portion $\ 0 \le x,y,z\le1 \ $ which is under the plane $\ x+y+z=2\ $ and above the plane $\ y+z=1\ $.
My solution
Let $D=\{x+y\le2 \ ,\ y\ge1\}$ we have \begin{equation} \begin{split} \iiint_R y\;dxdydz=&\iint_D y\;dxdy\int_{1-y}^{2-x-y}\;dz\\ =&\iint_D\ {y(1-x)}\;dxdy\\ =&\int_0^2(1-x)\;dx\int_1^{2-x}y\;dy\\ =&\frac{1}{2}\int_0^2(1-x)(4+x^2-4x-1)\;dx\\ =&\frac{2}{3} \end{split} \end{equation}
I'm not sure of the validity of my solution, could someone help me? Thanks in advance!
Definitely wrong: the intersection between the planes $x + y + z = 2$ and $y + z = 1$ is the same that the intersection of the planes $y + z = 1$ and $x = 1$ [$x = x + 1 - 1 = (x + (y + z)) - 1 = 2 - 1 = 1$], but $x = 1$ contains one of the faces of the unit cube, so the projection of the body on the $XY$ plane is the unit square, and the integral is $$ \int_0^1\int_0^1\int_{\max(0,1 - y)}^{\min(1,2 - x - y)}1\,dz\,dy\,dx = \int_0^1\int_0^1\int_{1-y}^{\min(1,2 - x - y)}1\,dz\,dy\,dx = \cdots $$ (why $\max(0,1 - y) = 1 - y$?). Now, $2 - x - y\le 1$ iff $x + y\ge 1$ and $$ \cdots = \int_0^1\int_0^{1-x}\int_{1-y}^1 1\,dz\,dy\,dx + \int_0^1\int_{1-x}^1\int_{1-y}^{2-x-y}1\,dz\,dy\,dx = \cdots $$