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October 16th, 2017, 07:13 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  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}^{1x} \int_{0}^{x+2y} (3+2x+2y2z) ~dz~dy~dx = 5/3 $ What did I do wrong? 
October 16th, 2017, 07:24 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I get 5/3 also just calculating the integral you wrote down. Maybe the question/integral was written down wrong?


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density, integral, shape, strange, triple 
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