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: 17


$6 \pi$

Work Step by Step

Since, the region $R$ is defined as follows: $R=${$(u,v) | 0 \leq v \leq 1-u, 0\leq u \leq 1$} $Jacobian (u,v)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 2&0\\0&3\end{vmatrix}=6$ $\iint_R x^2 dA=\int_{-1}^1 [\int_{-\sqrt{1-b^2}}^{\sqrt{1-b^2}} 24 u^2 da db$ Plug $a=r \cos \theta$ and $b=r \sin \theta$ $\iint_R x^2 dA=\int_{-1}^1 [\int_{-\sqrt{1-b^2}}^{\sqrt{1-b^2}} 24 a^2 da db=\int_0^{2 \pi} \int_0^1 24 (r \cos \theta)^2 r dr d \theta$ or, $\int_0^{2 \pi} 6 \cos^2 d \theta$ or, $= \int_0^{2 \pi} (3) (1+\cos 2 \theta) d \theta$ or, $=3 [\theta+\dfrac{\sin 2 \theta}{2}]_0^{2 \pi}$ Thus, $\iint_R x^2 dA=6 \pi$
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