Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.2 - Volumes Using Cylindrical Shels - Exercises 6.2 - Page 329: 9

Answer

We need to use the shell model as follows: $V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx$ $ \implies V= \int_0^{1} (2 \pi) \cdot (x)[2-x-x^2] dx$ Now, $V=2 \pi [x^2-\dfrac{x^3}{3}-\dfrac{x^4}{4}]_0^1$ or, $= 2 \pi \times (1-\dfrac{1}{3}-\dfrac{1}{4})$ or, $=\dfrac{5 \pi}{6}$

Work Step by Step

We need to use the shell model as follows: $V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx$ $ \implies V= \int_0^{2} (2 \pi) \cdot (x)[2-x-x^2] dx$ Now, $V=2 \pi [x^2-\dfrac{x^3}{3}-\dfrac{x^4}{4}]_0^1$ or, $= 2 \pi \times (1-\dfrac{1}{3}-\dfrac{1}{4})$ or, $=\dfrac{5 \pi}{6}$
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.