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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.

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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 $\int_2^0...$

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With reference to the following sketch

enter image description here

the set up should be

$$\int_0^2 dx \int_0^4 dz \int_z^{z+1}dy$$

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