Calculus: Early Transcendentals 9th Edition

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

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 456: 1

Answer

a) See image b) $\int_{0}^{3} \pi(x^4+10x^2+25) dx $ c) $\frac{1068\pi}{5}$

Work Step by Step

a) img b) $\\volume = \int\pi(y)^2$ apply $\ y=x^2+5$ to formula, $\ \pi(x^2+5)^2 = \pi(x^4+10x^2+25)$ shaded region is 0 to 3, so $\int_{0}^{3} \pi(x^4+10x^2+25) dx $ c) 1) Apply Sum Rule of Integration\(\displaystyle \pi \int _{0}^{3}x^{4}+10x^{2}+25~dx\) \(\displaystyle \pi (\int _{0}^{3}x^{4}~dx+\int _{0}^{3}10x^{2}~dx+\int _{0}^{3}25~dx)\) 2)Apply the Constant Multiple Rule of Integration \(\displaystyle \pi (\int _{0}^{3}x^{4}~dx+\int _{0}^{3}{10}x^{2}~dx+\int _{0}^{3}{25}~dx)\) \(\displaystyle \pi (\int _{0}^{3}x^{4}~dx+{10}\int _{0}^{3}x^{2}~dx+{25}\int _{0}^{3}1~dx)\) 3)Apply the Power Rule of Integration\(\displaystyle \pi (\int _{0}^{3}x^{4}~dx+10\int _{0}^{3}x^{2}~dx+25\int _{0}^{3}1~dx)\)\(\displaystyle \pi \left[\frac{x^{4+1}}{4+1}+10\cdot \frac{x^{2+1}}{2+1}+25\cdot \frac{x^{0+1}}{0+1}\right]_{0}^{3}\) 4)Simplify \(\displaystyle \pi \left[\frac{1}{5}x^{5}+\frac{10}{3}x^{3}+25x\right]_{0}^{3}\) 5)Substitute and subtract\(\displaystyle \pi \left[\frac{1}{5}x^{5}+\frac{10}{3}x^{3}+25x\right]_{0}^{3}\) \(\pi \left(\frac{1}{5}\cdot 3^{5}+\frac{10}{3}\cdot 3^{3}+25\cdot 3-\left(\frac{1}{5}\cdot 0^{5}+\frac{10}{3}\cdot 0^{3}+25\cdot 0\right)\right)\) 6)Simplify \(\pi \cdot \frac{1068}{5}\)
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