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: 28


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

Work Step by Step

Need to prove that $\iint_R f(x+y) dA=\int_0^1u f(u)$ $Jacobian(x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 1&1\\1& -1\end{vmatrix}=-2$ Then, we have $J(u,v)=|-\dfrac{1}{2}|=\dfrac{1}{2}$ Let us consider that $u=x+y$ and $v=x-y$ $\iint_R f(x+y) dA=(\dfrac{1}{2}) \int_{0}^{1} \int_{-u}^{u} f(u) dv du$ This implies that $\iint_R f(x+y) dA=\int_{0}^1 (\dfrac{1}{2}) [u-(-u)] f(u) du= \int_0^1(2)(\dfrac{1}{2})u f(u)=\int_0^1u f(u)$ Hence, our result is: $\iint_R f(x+y) dA=\int_0^1u f(u)$
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