Thomas' Calculus 13th Edition

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

Chapter 15: Multiple Integrals - Additional and Advanced Exercises - Page 935: 2


$\dfrac{52 \pi}{3}$ or, $54.45427$

Work Step by Step

We are given the conditions $9=25-x^2-y^2 \implies x^2+y^2=16$ which is the projection of the volume of water onto the $xy$-plane. The rectangular coordinates can be converted into polar coordinates as : $x =r \cos \theta$ and $y=r \sin \theta$ and $r^2=x^2+y^2$ So, we can write the volume for triple integration as: $V=\int_{0}^{2 \pi} \int_{0}^{4} \int_{-\sqrt {25-r^2}}^{-3} \ dz (r \ dr )\ d \theta$ or, $=\int_{0}^{2 \pi} [\int_{0}^{4} (r \sqrt {25-r^2} \ dr -\int_{0}^{4} 3r \ dr ]\ d \theta$ or, $=\int_{0}^{2 \pi} [(\dfrac{-1}{3} (25-r^2)^{3/2})_0^4-[\dfrac{3r^2}{2}]_{0}^{4} ] \ d \theta$ or, $=\int_{0}^{2 \pi} [\dfrac{98}{3}-\dfrac{3}{2}(4^2-0) \ d \theta$ or, $=\int_{0}^{2 \pi} [\dfrac{98}{3}-24] \ d \theta$ or, $=\int_0^{2 \pi} \dfrac{26}{3} \ d \theta$ or, $=\dfrac{52 \pi}{3}$ or, $=54.45427$
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