Thomas' Calculus 13th Edition

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

Chapter 6: Applications of Definite Integrals - Section 6.2 - Volumes Using Cylindrical Shels - Exercises 6.2 - Page 330: 27

Answer

$a)$ $\frac{6\pi}{5}$ $b)$ $\frac{4\pi}{5}$ $c)$ $2\pi$ $d)$ $2\pi$

Work Step by Step

a) the x axis $V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(y)[12(y^{2}-y^{3}$)]$dy$ $V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$[(y^{3}-y^{4}$)]$dy$ $V$ = $24\pi$$(\frac{y^{4}}{4}-\frac{y^{5}}{5})$ $|_{{\,0}}^{{\,1}}$ $V$ = $24\pi$$[(\frac{1}{4}-\frac{1}{5})-(0-0)]$ $V$ = $\frac{6\pi}{5}$ b) the line y=1 $V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(1-y)[12(y^{2}-y^{3}$)]$dy$ $V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(y^{2}-2y^{3}+y^{4}$)$dy$ $V$ = $24\pi$$(\frac{y^{3}}{3}-\frac{y^{4}}{2}+\frac{y^{5}}{5})$ $|_{{\,0}}^{{\,1}}$ $V$ = $24\pi$$[(\frac{1}{3}-\frac{1}{2}+\frac{1}{5})-(0-0+0)]$ $V$ = $\frac{4\pi}{5}$ c) line y=$\frac{8}{5}$ $V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(\frac{8}{5}-y)[12(y^{2}-y^{3}$)]$dy$ $V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(\frac{8}{5}y^{2}-\frac{13}{5}y^{3}+y^{4}$)$dy$ $V$ = $24\pi$$(\frac{8}{15}y^{3}-\frac{13}{20}y^{4}+\frac{1}{5}y^{5}$)$dy$$|_{{\,0}}^{{\,1}}$ $V$ = $24\pi$$[(\frac{8}{15}-\frac{13}{20}+\frac{1}{5})-(0-0+0)]$ $V$ = $2\pi$ d) the line y=-$\frac{2}{5}$ $V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(\frac{2}{5}+y)[12(y^{2}-y^{3}$)]$dy$ $V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(\frac{2}{5}y^{2}+\frac{3}{5}y^{3}-y^{4}$)$dy$ $V$ = $24\pi$$(\frac{2}{15}y^{3}+\frac{3}{20}y^{4}-\frac{1}{5}y^{5}$)$dy$$|_{{\,0}}^{{\,1}}$ $V$ = $24\pi$$[(\frac{2}{15}+\frac{3}{20}-\frac{1}{5})-(0+0-0)$)]$dy$ $V$ = $2\pi$
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