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October 16th, 2017, 08:13 AM   #1
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Triple integral density of a strange shape.

x = 0
y = 0
z = 0
x + y = 1
z = x + 2y

Density(x, y, z) = 3 + 2x + 2y - 2z

Answer: 1/6

My Solution (not correct)
$\displaystyle
\int_{0}^{1} \int_{0}^{1-x} \int_{0}^{x+2y} (3+2x+2y-2z) ~dz~dy~dx = 5/3
$

What did I do wrong?
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October 16th, 2017, 08:24 AM   #2
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I get 5/3 also just calculating the integral you wrote down. Maybe the question/integral was written down wrong?
Thanks from romsek and zollen
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