Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.4 Applications of Double Integrals - 15.4 Exercises - Page 1064: 3

Answer

$m$ = $42k$ $\left(2,\frac{85}{28}\right)$

Work Step by Step

$m$ = $\int_1^3\int_1^4ky^2dydx$ $m$ = $\frac{1}{3}\int_1^3ky^3|_1^4dx$ $m$ = $\int_1^3{21k}dx$ $m$ = $21kx|_1^3$ $m$ = $42k$ $x̄$ = $\frac{1}{m}\int_1^3\int_1^4kxy^2dydx$ $x̄$ = $\frac{1}{42k}\int_1^3\int_1^4kxy^2dydx$ $x̄$ = $\frac{1}{42}\int_1^3{\frac{xy^3}{3}}|_1^4dx$ $x̄$ = $\frac{1}{42}\int_1^3{21x}dx$ $x̄$ = $\frac{1}{2}\int_1^3{x}dx$ $x̄$ = $\frac{1}{4}[x^2]|_1^3$ $x̄$ = $2$ $ȳ$ = $\frac{1}{m}\int_1^3\int_1^4ky^3dydx$ $ȳ$ = $\frac{1}{42k}\int_1^3\int_1^4ky^3dydx$ $ȳ$ = $\frac{1}{42}\int_1^3[\frac{y^4}{4}]_1^4dx$ $ȳ$ = $\frac{1}{42}\int_1^3{\frac{255}{4}}dx$ $ȳ$ = $\frac{255}{168}[x]_1^3$ $ȳ$ = $\frac{85}{28}$
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