Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 7 - Applications of Integration - 7.2 Exercises: 15

Answer

$$V=18\pi$$

Work Step by Step

$V=\pi \int_0^3(4-x)^2-(4-3)^2 dx$ $V=\pi\int_0^3[(x^2-8x+16)-(1)^2] dx$ $V=\pi\int_0^3(x^2-8x+15) dx$ $V=\pi [\frac{x^3}{3}-4x^2+15x)]_0^3$ $V=\pi[(\frac{3^3}{3}-4(3)^2+15(3))-(\frac{(0)^3}{3}-4(0)+15(0))]$ $V=\pi[(9-36+45)-(0)]$ $V=18\pi$
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