Calculus: Early Transcendentals 9th Edition

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

Chapter 15 - Review - Exercises - Page 1119: 38

Answer

$$12 \pi$$

Work Step by Step

$$Volume=\iint_{x^2+y^2 \leq 4} \int_{0}^{3-y} dz dA\\=\iint_{x^2+y^2 \leq 4} (3-y) dA \\=3 \iint_{x^2+y^2 \leq 4} dA - \iint_{x^2+y^2 \leq 4} y dA$$ The inner integral is the integral of an odd continuous function in a symmetric interval; this implies that $\iint_{x^2+y^2 \leq 4} y dA=0$ and $$Volume=(3) \times (\pi) \times (2)^2 -0=12 \pi$$
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