Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Review - Exercises - Page 691: 42

Answer

$\frac{1}{2} \pi [\sinh(6)+6]$

Work Step by Step

Given: $x=2+3t$ $\implies$ $\frac{dx}{dt}=3$ $y=\cosh (3t)$ $\implies$ $y=3 \sinh(3t)$ To calculate the area of the surface, we will use the equation: $A=2 \pi \int_0^1 y(t) \sqrt{(dx/dt)^2+(dy/dt)^2}dt$ $=2 \pi \int_0^1 [\cosh (3t)] \sqrt{3^2+(3 \sinh(3t))^2}dt$ $=6 \pi \int_0^1 [\cosh (3t)] \sqrt{1+( \sinh^2(3t))}dt$ Since, $\cosh (3t) \gt 0$ Thus, $=6 \pi \int_0^1 [\cosh^2 (3t)] dt$ $=6 \pi \int_0^1 (\dfrac{e^{3t}+e^{-3t}}{2})^2 dt$ $=\frac{3}{2} \pi [ \dfrac{e^{6t}}{6}+2t+\frac{e^{-6t}}{-6}]_0^1$ $=\frac{1}{2} \pi [\sinh(6)+6]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.