# Triple Integral bounds question

For the solid bounded by $$x=0 , x=2, z=y, z=y-1, z=0, z=4$$ I am looking at the $$yz-plane$$ and setting up a triple integral. I get $$\int_{0}^{4} \int_{z+1}^{z} \int_{0}^{2} dxdydz$$ However, I was wondering if I did this correctly because $$z+1$$ is greater than $$z$$, but when I drew the projection on the yz plane, it looked like z=y-1 is under z=y. Any help appreciated.

You need to switch the two limits. In a way it is similar to saying if the solid is bounded by $$x=2, x=0, ...$$ You would not set up the integral as $$\int_2^0...$$
$$\int_0^2 dx \int_0^4 dz \int_z^{z+1}dy$$