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 October 16th, 2017, 08:13 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) $\displaystyle \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{x+2y} (3+2x+2y-2z) ~dz~dy~dx = 5/3$ What did I do wrong? October 16th, 2017, 08:24 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,166 Thanks: 738 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? Thanks from romsek and zollen Tags density, integral, shape, strange, triple Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 2 May 29th, 2017 04:11 AM sg555 Calculus 3 April 11th, 2016 11:29 PM karans88 Probability and Statistics 7 November 13th, 2014 04:09 AM Suliman Calculus 1 September 21st, 2013 12:33 AM ZardoZ Real Analysis 6 August 10th, 2011 04:18 AM

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