Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.7 - Triple Integrals in Cylindrical and Spherical Coordinates - Exercises 15.7 - Page 921: 60

Answer

$32\pi $

Work Step by Step

We know that the paraboloid will intersect the xy-plane when: $9-x^2-y^2=0$ Solving, we get: $ x^2+y^2=9$ Now, calculate the volume: $ V=4\int^{\pi/2}_0 \int^3_1 \int^{9-r^2}_0$ dz r dr $ d\theta $ =$4\int^{\pi/2}_0 \int^3_1 (9r-r^3)dr $ $ d\theta $ =$4\int^{\pi/2}_0 [\frac{9r^2}{2}-\frac{r^4}{4}]^{3}_1 d\theta $ =$4\int^{\pi/2}_0 (\frac{81}{4}-\frac{17}{4})d\theta $ =$64 \int^{\pi/2}_0 d\theta $ =$32\pi $
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