# Thread: Volume using Triple Integrals Setup?

1. ## Volume using Triple Integrals Setup?

Would dz be from 0 to 3, dx from 0 to 4, and dy from z to (3-z)/2 be correct? for 16a?

2. ## Re: Volume using Triple Integrals Setup?

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