Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.5 Area Of A Surface Of Revolution - Exercises Set 5.5 - Page 380: 1

Answer

$$S = 35\sqrt 2 \pi $$

Work Step by Step

$$\eqalign{ & y = 7x,\,\,\,\,0 \leqslant x \leqslant 1 \cr & {\text{Calculate }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = 7 \cr & {\text{Use }}S = \int_a^b {2\pi y\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr & S = \int_0^1 {2\pi \left( {7x} \right)\sqrt {1 + {{\left( 7 \right)}^2}} } dx \cr & S = \int_0^1 {14\pi x\sqrt {50} } dx \cr & S = 70\sqrt 2 \pi \int_0^1 x dx \cr & {\text{Integrate}} \cr & S = 70\sqrt 2 \pi \left[ {\frac{1}{2}{x^2}} \right]_0^1 \cr & S = 35\sqrt 2 \pi \left[ {{x^2}} \right]_0^1 \cr & S = 35\sqrt 2 \pi \cr} $$
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.