Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.9 Change of Variables in Multiple Integrals - 15.9 Exercises - Page 1100: 20

Answer

$\dfrac{3}{4}$

Work Step by Step

Since, $|J| =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{2u}{v}&-\dfrac{v}{u^2}\\\dfrac{-u^2}{v^2}& \dfrac{1}{u}\end{vmatrix}=v^{-1}$ $\iint_R y^2 dA=\int_1^{2} \int_{1}^{2}(\dfrac{v}{u})^2(v^{-1}) du dv$ and $\iint_R y^2 dA=\int_1^2 v dv \int_{1}^{2} u^{-2} du$ Therefore, $\iint_R y^2 dA=[\dfrac{v^2}{2}]_1^2[\dfrac{-1}{u}]_1^2=\dfrac{3}{4}$
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