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: 18


$\dfrac{4 \sqrt 3 \pi}{3}$

Work Step by Step

Here, we have: $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \sqrt 2&-\sqrt{\dfrac{2}{3}}\\\sqrt 2&\sqrt{\dfrac{2}{3}}\end{vmatrix}=\dfrac{4 \sqrt 3}{3}$ $\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)\dfrac{4 \sqrt 3}{3} dA$ Here, $u^2+v^2 \leq 1$ Plug $u =r \cos \theta$ and $v=r \sin \theta$ $\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)\dfrac{4 \sqrt 3}{3} dA=\int_0^{2 \pi} \int_0^1 2r^2 (\dfrac{4 \sqrt 3}{3}) r dr d \theta$ Thus, $\iint_R (x^2-xy+y^2) dA=\dfrac{8 \sqrt 3}{3}\int_0^{2 \pi} \dfrac{1}{4} r^4|_0^1 d \theta=\dfrac{4 \sqrt 3 \pi}{3}$
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