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: 26

Answer

$\dfrac{\pi}{24} (1-\cos 1)$

Work Step by Step

$Jacobian, J (x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{1}{3}&0\\0&\dfrac{1}{2}\end{vmatrix}=\dfrac{1}{6}$ Let us consider that $u=3x$ and $v=2y$ $\iint_R \sin(9x^2+4y^2) dx dy= \iint_{R} \sin (u^2+v^2) (\dfrac{1}{6}) du dv$ or, $=(\dfrac{1}{6}) \int_0^{\pi/2} \int_0^1 \sin (r^2) r dr d \theta$ or, $ =(\dfrac{-1}{12}) (\pi/2) |\cos (r^2)|_0^1$ or, $=\dfrac{-1}{12}(\cos (1)-1)$ or, $=\dfrac{\pi}{24} (1-\cos 1)$
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