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Thread: Volume using Triple Integrals Setup?

  1. #1
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    Volume using Triple Integrals Setup?

    Volume using Triple Integrals Setup?-mathform.png

    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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  2. #2
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    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
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