Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 8 - Section 8.3 - Applications to Physics and Engineering. - 8.3 Exercises - Page 585: 28

Answer

$\left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{\ln 2}},\frac{1}{{4\ln 2}}} \right)$

Work Step by Step

$$\eqalign{ & {\text{From the given graph}}{\text{, we can estimate the centroid at}} \cr & \left( {1.5,0.3} \right) \cr & \cr & y = \frac{1}{x},{\text{ }}1 \leqslant x \leqslant 2 \cr & {\text{Let }}f\left( x \right) = \frac{1}{x}{\text{ and }}g\left( x \right) = 0 \cr & \cr & {\text{*The mass of the lamina is }} \cr & m = \rho \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & m = \rho \int_1^2 {\left( {\frac{1}{x}} \right)} dx \cr & m = \rho \left[ {\ln x} \right]_1^2 \cr & m = \left( {\ln 2} \right)\rho \cr & \cr & *{\text{The moment about the }}x{\text{ - axis is}} \cr & {M_x} = \rho \int_a^b {\left[ {\frac{{f\left( x \right) + g\left( x \right)}}{2}} \right]} \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_x} = \rho \int_1^2 {\left[ {\frac{{\frac{1}{x}}}{2}} \right]} \left[ {\frac{1}{x}} \right]dx \cr & {M_x} = \frac{\rho }{2}\int_1^2 {\frac{1}{{{x^2}}}} dx \cr & {M_x} = \frac{\rho }{2}\left[ { - \frac{1}{x}} \right]_1^2 \cr & {M_x} = \frac{\rho }{2}\left[ { - \frac{1}{2} + \frac{1}{1}} \right] \cr & {M_x} = \frac{\rho }{4} \cr & \cr & *{\text{The moment about the }}y{\text{ - axis is}} \cr & {M_y} = \rho \int_a^b x \left[ {f\left( x \right) - g\left( x \right)} \right]dx \cr & {M_y} = \rho \int_0^2 x \left[ {\frac{1}{x}} \right]dx \cr & {M_y} = \rho \int_1^2 {dx} \cr & {M_y} = \rho \left[ x \right]_1^2 \cr & {M_y} = \rho \left( {2 - 1} \right) \cr & {M_y} = \rho \cr & \cr & *{\text{The coordinates }}\left( {\overline x ,\overline y } \right){\text{ of the centroid are:}} \cr & \overline x = \frac{{{M_y}}}{m} = \frac{\rho }{{\left( {\ln 2} \right)\rho }} = \frac{1}{{\ln 2}} \approx 1.44 \cr & \overline y = \frac{{{M_x}}}{m} = \frac{{\frac{\rho }{4}}}{{\left( {\ln 2} \right)\rho }} = \frac{1}{{4\ln 2}} \approx 0.36 \cr & \left( {\overline x ,\overline y } \right) = \left( {\frac{1}{{\ln 2}},\frac{1}{{4\ln 2}}} \right) \cr} $$
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.