Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 656: 57

Answer

$$ x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq \pi / 2 $$ The surface area, which is obtained by rotating the given curve about the x-axis, is $$ S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi t \cos t \sqrt{t^{2}+1} d t \approx 4.7394 $$

Work Step by Step

$$ x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq \pi / 2 $$ then $$ d x / d t=t \cos t+\sin t , \quad d y / d t =-t \sin t+\cos t, $$ so $$ (d x / d t)^{2}+(d y / d t)^{2}=t^{2} \cos ^{2} t+2 t \sin t \cos t+\sin ^{2} t+t^{2} \sin ^{2} t-2 t \sin t \cos t+\cos ^{2} t \\ \quad \quad \quad \quad\quad\quad=t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)+\sin ^{2} t+\cos ^{2} t=t^{2}+1 $$ $$ \begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=t^{2} \cos ^{2} t+2 t \sin t \cos t+\sin ^{2} t+t+\\ & \quad \quad +^{2} \sin ^{2} t-2 t \sin t \cos t+\cos ^{2} t \\ &=t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)+\sin ^{2} t+\cos ^{2} t \\ &=t^{2}+1 \end{aligned} $$ Thus the surface area, which is obtained by rotating the given curve about the x-axis, is $$ S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi t \cos t \sqrt{t^{2}+1} d t \approx 4.7394 $$
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