Calculus: Early Transcendentals 8th Edition

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

Chapter 15 - Section 15.2 - Double Integrals over General Regions - 15.2 Exercise - Page 1010: 66

Answer

$\dfrac{2}{3} \pi R^3$

Work Step by Step

We will re-write $z=\sqrt {R^2-x^2-y^2} \\ z^2=R^2-x^2-y^2 \\ x^2+y^2+z^2 =R^2$ The equation of an upper half of a hemi-sphere is: $x^2+y^2+z^2 =R^2$ when $z \geq 0$ So, we can find the volume as follows: $Volume =\dfrac{1}{2} \times \dfrac{4}{3} \pi R^3 =\dfrac{2}{3} \pi R^3$
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