$\iiint (y^2+z) dxdydz$ under region $x+y+z=2, x=2, y=1, z=y$ I'm having trouble trying to find the interval.
I tried mapping it to $u=x+y+z$,  $v=x$,  $w=y$ to get
$$u=2$$
$$v=2$$
$$w=1$$
$$u-v-w=w$$
Is this in the right direction?
 A: Your region is the union of the two regions $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, z\leq y\leq 1, 0\leq z\leq 1\}$ and $\{(x,y,z)\in\mathbb{R}^3:2-y-z\leq x\leq 2, -z\leq y\leq 1, -1\leq z\leq 0\}$. 
This can be determined by sketching the region in the $(y,z)$ plane.
Your integral is
$$\int_0^1\int_z^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz + \int_{-1}^0\int_{-z}^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz$$
The first integral can be calculated as follows:
$$\begin{align}\int_0^1\int_z^1\int_{2-y-z}^2(y^2+z)\,dx\,dy\,dz &= \int_0^1\int_z^1x(y^2+z)|_{2-y-z}^2\,dy\,dz\\ 
&= \int_0^1\int_z^1(y^3+yz+zy^2+z^2)\,dy\,dz\\
&=\int_0^1 \left(\frac{y^4}{4}+\frac{y^2}{2}+\frac{zy^3}{3}+z^2y\right)\Bigg|_z^1\,dz\\
&=\int_0^1 \left(\frac{1}{4}+\frac{5}{6}z+z^2-\frac{3}{2}z^3-\frac{7}{12}z^4\right)\,dz\\
&=\left(\frac{1}{4}z+\frac{5}{12}z^2+\frac{1}{3}z^3-\frac{3}{8}z^4-\frac{7}{60}z^5\right)\Bigg|_0^1\\
&=\frac{61}{120}
\end{align}$$
The second integral can be calculated similarly, and is equal to $\frac{7}{120}$. Therefore, the answer is $\frac{68}{120}=\frac{17}{30}$.
