Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - 15.6 Triple Integrals - 15.6 Exercises - Page 1077: 8



Work Step by Step

$ \int_{0}^1 \int_{0}^{1} \int_{0}^{2-x^2-y^2} xye^z dz dy dx= \int_{0}^1 \int_{0}^{1} [xye^z]_{0}^{2-x^2-y^2} dy dx$ This implies that $\int_{0}^1 \int_{0}^{1} [xye^{2-x^2-y^2}-xye^0] dy dx=(e^2) \int_{0}^{1} ye^{-y^2} dy \int_0^1 xe^{-x^2} dx-(\dfrac{x^2}{2})_0^1[\dfrac{y^2}{2}]_0^1 $ Now consider $p=-x^2; dp=(2x) dx$ $ (e^2) [ \int_0^1 xe^{-x^2} dx-(\dfrac{x^2}{2})]-(\dfrac{1}{2})(\dfrac{1}{2}) = (e^2) [(-\dfrac{1}{2}) \int_0^{-1}e^p dp]^2-\dfrac{1}{4} = (e^2) [\dfrac{1-(1/e)}{2}]^2-(1/4)$ Hence, $\int_{0}^1 \int_{0}^{1} \int_{0}^{2-x^2-y^2} xye^z dz dy dx =\dfrac{(e-1)^2-1}{4}=\dfrac{e^2-2e}{4}$
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