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 657: 66

Answer

$\pi e^2-6\pi+2\pi e\approx 21.443$

Work Step by Step

$\begin{gathered}\frac{\mathrm{d} x}{\mathrm{~d} t}=e^t-1 \\ \frac{\mathrm{d} y}{\mathrm{~d} t}=2 e^{t / 2} \\ S=\int_0^1 2 \pi\left(e^t-t\right) \sqrt{\left(e^t-1\right)^2+\left(2 e^{t / 2}\right)^2} d t \\ =\int_0^1 2 \pi\left(e^t-t\right) \sqrt{e^{2 t}-2 e^t+1+4 e^t} d t \\ =\int_0^1 2 \pi\left(e^t-t\right) \sqrt{e^{2 t}+2 e^t+1} d t \\ =\int_0^1 2 \pi\left(e^t-t\right) \sqrt{\left(e^t+1\right)^2} d t \\ =\int_0^1 2 \pi\left(e^t-t\right)\left(e^t+1\right) d t \\ =\int_0^1 2 \pi\left(e^{2 t}-t e^t+e^t-t\right) d t \\ I=\int t e^t d t \\ u=t, d v=e^t \\ I=t e^t-\int e^t d t=t e^t-e^t \\ S=\left.2 \pi\left[\frac{e^{2 t}}{2}-t e^t+2 e^t-\frac{t^2}{2}\right]\right|_{t=0} ^{t=1}=\pi e^2-6\pi+2\pi e=21.443\end{gathered}$
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