Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 446: 12

Answer

$\displaystyle{V=\frac{471\pi }{14}}\\ $

Work Step by Step

$\displaystyle{A\left(x\right)=\pi \left(x^3+3\right)^2-\pi \left(1+3\right)^2}\\ \displaystyle{A\left(x\right)=\pi \left(7-6x^3-x^6\right)}\\$ $\displaystyle{V=\int_{1}^2A\left(x\right)\ dx}\\ \displaystyle{V=\int_{1}^2\pi \left(7-6x^3-x^6\right)\ dx}\\ \displaystyle{V=\pi \int_{1}^27-6x^3-x^6\ dx}\\ \displaystyle{V=\pi\left[7x-\frac{3}{2}x^4-\frac{1}{7}x^7\right]_{1}^2}\\ \displaystyle{V=\pi\left(\left(7(2)-\frac{3}{2}(2)^4-\frac{1}{7}(2)^7\right)-\left(7(1)-\frac{3}{2}(1)^4-\frac{1}{7}(1)^7\right)\right)}\\ \displaystyle{V=\frac{471\pi }{14}}\\ $
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