Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 731: 41


$\dfrac{471,295 \pi}{1024}$

Work Step by Step

Given: $x=4 \sqrt t$ $\implies$ $\frac{dx}{dt}=\frac{2}{ \sqrt t}$ $y=\frac{t^3}{3}+\frac{1}{2t^2}$ $\implies$ $y=t^2-\frac{1}{t^3}$ To calculate the area of surface, we will use the equation: $A=2 \pi \int_1^4 y(t) \sqrt{(dx/dt)^2+(dy/dt)^2}dt$ $=2 \pi \int_1^4 (\frac{t^3}{3}+\frac{1}{2t^2}) \sqrt{(\frac{2}{ \sqrt t})^2+(t^2-\frac{1}{t^3})^2}dt$ $=2 \pi \int_1^4 [\frac{t^3}{3}+\frac{1}{2} +\frac{1}{3}+\frac{1}{2t^5}]dt$ $=2 \pi \int_1^4 [\frac{t^5}{3}+\frac{5}{6} +\frac{1}{2t^5}]dt$ $=2 \pi[\frac{t^6}{18}+\frac{5t}{6} -\frac{1}{8t^4}]_1^4$ $=\dfrac{471,295 \pi}{1024}$
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.