Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - 15.3 Exercises - Page 1019: 10

Answer

$$ \iint \limits_{D} x^3d A=\frac{3e^4}{16}+\frac{1}{16}$$

Work Step by Step

Given $$ \iint \limits_{D} x^3d A$$ $$D=\{(x, y) |1\leq x \leq e,0\leq y \leq \ln x\}$$ So, we have \begin{aligned} A&=\iint \limits_{D} x^3 \ d A\\ &=\int_{0}^{e} \int_{0}^{\ln x} x^3 \ d y \ d x\\ &=\int_{0}^{e} x^3 \left[ y\right]_{0}^{\ln x} \ d x\\& = \int_{0}^{e}x^3 \ln x \ d x\\ \end{aligned} So, by the partition technique, let $$u=ln x \Rightarrow du= \frac{1}{x}dx$$ $$dv=x^3 \ dx \Rightarrow v=\frac{x^4}{4} $$ So, we get \begin{aligned} A &=\left[uv \right]_{1}^{e}- \int_{1}^{e} vdu\\ & =\left[\frac{x^4 \ln x}{4} \right]_{1}^{e}- \int_{1}^{e}\frac{x^3}{4} \ d x\\ &=\left[\frac{x^4 \ln x}{4} \right]_{1}^{e}-\left[ \frac{x^4}{16} \right]_{1}^{e} \\ &=\left[\frac{e^4 \ln e}{4} -0 \right] -\left[\frac{e^4}{16} -\frac{1}{16} \right] \\ &= \frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}\\ &=\frac{3e^4}{16}+\frac{1}{16} \end{aligned}
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