Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.9 - Change of Variables in Multiple Integrals - 15.9 Exercise - Page 1060: 28

Answer

$\iint_R f{x+y} dA=\int_0^1u f(u)$

Work Step by Step

Consider $u=x+y$ and $v=x-y$ Here, we have: $Jacobian (x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 1&1\\1& -1\end{vmatrix}=(1) \cdot (-1) -(1)(1)=-2$ and $J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$ Now, $\iint_R f{x+y} dA=(\dfrac{1}{2}) \int_{0}^{1} \int_{-u}^{u} f(u) dv du$ or, $=(\dfrac{1}{2}) \int_{0}^1 [u-(-u)] f(u) du$ Thus, we have $\iint_R f{x+y} dA=(\dfrac{1}{2}) \int_0^1(2u) f(u)=\int_0^1u f(u)$ Hence, it has been prove that $\iint_R f{x+y} dA=\int_0^1u f(u)$
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