The integral I've formed here is: $$\int_1^2 \int_0^1 \int^{3y}_y 1 dzdydx$$
Which evaluates to one. Not sure if I'm thinking of the region correctly, does this integral make sense for this region?
Also, a very similar question: Find the volume of the region in the first octant bounded by the coordinate planes and the surface $++= 2$. Would the integral for this region be $$\int_0^2 \int_0^2 \int^{x+y-2}_0 1 dzdydx?$$
I feel as though if I've got just one of those right then I got both right, because I used the same thinking...any help is appreciated!
The integral expression
$$\int_1^2 \int_0^1 \int^{3y}_y dzdydx$$
is correct. The volume is indeed 1, which can be verified as follows. Note the cross section in the $yz$-plane is a triangle with base 2 and height 1, which gives an area 1. Then, multiple it with the thickness 1 in the $x$-direction to get the volume 1.
The second volume integral should be
$$\int_0^2 \int_0^{2-x} \int^{2-x-y}_0 dzdydx=\frac 43$$
Note that it is a cone with a right triangle base of area 2 and height 2. Thus, its volume is $\frac 13 \times 2\times 2= \frac 43$.