Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.6 - Triple Integrals - 15.6 Exercise - Page 1094: 42

Answer

$4\pi$

Work Step by Step

We can write the equation of a region as: $\iiint_{B} z^3+\sin y +3 \ dV=\iiint_{B} z^3+\sin y \ dV+\iiint_{B} 3 \ dV(1)$ But, $\iiint_{B} z^3+\sin y \ dV=0$ because the region inside the sphere is symmetrical, so the integral of an odd function will be zero. Thus, equation (1) becomes: $\iiint_{B} z^3+\sin y +3 \ dV=\iiint_{B} 3 \ dV = 3V$ The volume $V$ of the sphere is: $V=\dfrac{4}{3}\pi r^3$ So, $\iiint_{B} z^3+\sin y +3 \ dV=3 \times \dfrac{4}{3}\pi r^3=4 \pi r^3=4 \pi (1)^3=4\pi$ $[\because Radius =1]$
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