Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.6 Double Integrals - 9.6 Exercises - Page 514: 29



Work Step by Step

We are given $z=8x+4y+10$ with $-1\leq x \leq1, 0\lt y \lt 3$ Replacing the limits for the region R and $f(x,y)=z$ $V=\int^{3}_{0}\int^{1}_{-1}(8x+4y+10)dxdy$ $V=\int^{3}_{0}[\int^{1}_{-1}(8x+4y+10)dx]dy$ solve the inner integral, treating y as a constant $\int^{1}_{-1}(8x+4y+10)dx$ integrate $=[4x^{2}+4xy+10x]^{1}_{-1}=4+4y+10-(4-4y-10)=8y+20$ $V=\int^{3}_{0}[\int^{1}_{-1}(8x+4y+10)dx]dy$ $V=\int^{3}_{0}(8y+20)dy$ integrating $V=4y^{2}+20y|^{3}_{0}=4(3)^{2}+20(3)-(4.0+20.0)=96$
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