Evaluate $\iiint dx\,dy\,dz$ betweem $x=0,y=0,z=0, x+y+2z=2$ 
I need to evaluate $$\iiint dx\,dy\,dz$$
the volume between $x=0,y=0,z=0, x+y+2z=2$

I am stuck about choosing the limits of integration, I think that the limits should be:
$$\int_0^{(2-x-y)/2}\int_0^{2-2z-x}\int_{0}^{2-y-2z}\,dx\,dy\,dz$$
If this is not correct, can you explain WHY please? Thank you.

 A: $x+y+2z=2$ is a plane which cut the axis $x,y$ and $z$ in $x=2,y=2$ and $z=1$ respectively. I draw a figure in which you can find out why the limits of $x$ and $y$ is chosen properly:

As you see we have $x\in [0..2]$ and of course $y\in [0..2-x]$. Note that above region is on $z=0$ so $$x+y+2\times 0=2$$ Obviously $z\in [0..\frac{2-x-y}{2}]$
A: See putting $x=0,y=0$ the plane cuts the $Z$ axis at $1$ . Similarly putting $x=0$, the plane cuts the $Y$ axis at $2-2z$ So the integration would be$$\int _0^1\int_0^{2-2z}\int _0^{2-y-2z}dxdydz$$
A: It's very useful to make a sketch. The plane $x+y+2z=2$, together with the coordinate planes, bound a region in the first octant.
In the $xy$-plane, that plane (set $z=0$ to get $x+y=2$) reduces to the line segment connecting $(2,0,0)$ to $(0,2,0)$.
Similarly, letting $x=0$ to get the line segment in the $yz$-plane and letting $y=0$ to get the line segment in the $xz$-plane, you get the pieces connecting $(0,2,0)$ to $(0,0,1)$ and $(2,0,0)$ to $(0,0,1)$. Sketching this should give you a good idea of the region you want to find the volume of.
In this particular case, you can choose the order of integration freely as it will be possible to integrate in all possible (6) orders and the difficulty is roughly the same. This is not necessarily the case for other regions.
If you choose $x$ as the outer integration variable, then clearly $x:0\to2$. For any $x \in [0,2]$, look in the $xy$-plane where the line segment lies on $x+y=2$ to solve $y=2-x$. Clearly $y$ starts at the $x$-axis so $y:0 \to 2-x$; you now have the limits describing the projection of the region onto the $xy$-plane which is a triangle, but that should be clear from your sketch.
Lastly, for any $(x,y)$ in this triangle, you need to integrate from bottom (clearly the $xy$-plane, so $z=0$) to the top, which is the plane given by $x+y+2z=2$, solving for $z$ yields its limits: $z : 0 \to 2-x/2-y/2$.
It's a good exercise to try and determine the limits if you would choose a different order of integration.
