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October 27th, 2017, 07:35 AM   #1
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Triple Integral of a complex geometric shape

Boundaries:
z = 0
z = x + 2y
z = 4 - x - 3y
z = -y

Answer:
$\displaystyle
\int_{0}^{4} \int_{-z}^{4-2z} \int_{z-2y}^{4-z-3y} ~dx~dy~dz = \frac {32}{3}

$

I would be much appreciated if anyone show me the steps of setting the above boundaries...
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