Would dz be from 0 to 3, dx from 0 to 4, and dy from z to (3-z)/2 be correct? for 16a?
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the integral will be $\displaystyle \int_0^1 \int_0^4 \int_z^{\frac{3-z}{2}}~dy~dx~dz$ look at where $z=y$ and $z = 3-2y$ intersect