Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.6 - Triple Integrals - 15.6 Exercise - Page 1037: 11


$\dfrac{9 \pi}{8}$

Work Step by Step

Consider $I=\iiint_E \dfrac{z}{x^2+z^2} dV$ $ I=\int_{1}^{4} \int_{y}^{4} \int_{0}^{z}\dfrac{z}{x^2+z^2} dx dzdy=\int_{1}^{4} \int_{y}^{4} [(z)\dfrac{1}{z}\arctan(\dfrac{x}{z}]_{0}^{z} dzdy$ or, $=\int_{1}^{4} \int_{y}^{4} \dfrac{\pi}{4} dzdy$ or, $= \int_{1}^{4} [\dfrac{\pi z}{4} ]_y^4dy$ or, $= \int_{1}^{4} [\pi-\dfrac{\pi y}{4} ] dy$ or, $=[\pi y-\dfrac{\pi}{8}y^2]_1^4$ or, $=\dfrac{9 \pi}{8}$
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