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 6.6 - Moments and Centers of Mass - Exercises 6.6 - Page 361: 29

Answer

$\overline {x} =\dfrac{6}{5}; \overline {y}=\dfrac{8}{7}$

Work Step by Step

We have $\overline {x}=\dfrac{3}{4} \int_{0}^{2} x \cdot (2x^2-x^3) dx $ or, $\overline {x} =\dfrac{3}{4} \times [ \dfrac{x^4}{2}-\dfrac{x^5}{5}]_{0}^{2} $ or, $\overline {x}= \dfrac{6}{5}$ and $\overline {y}=\dfrac{3}{4} \int_{0}^{2} \dfrac{x^3}{2}(2x^2-x^3) dx$ or, $\overline {y}=(\dfrac{3}{8}) \times [\dfrac{x^6}{3}-\dfrac{x^7}{7}]_{0}^{2} $ or, $\overline {y}=\dfrac{8}{7}$
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.