Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 59



Work Step by Step

Consider $V= \int_{-4}^{1} \int_{3x}^{4-x^2} (x+4) dy dx$ or, $= \int_{-4}^{1} [x y+4y]_{3x}^{4-x^2} dx$ or, $= \int_{-4}^{1} [4x-x^3+16-4x^2]-[3x^2+12x]dx$ or, $= \int_{-4}^{1} [-8x-x^3+16-7x^2]dx$ or, $= [-4x^2-\dfrac{x^4}{4}+16x-\dfrac{7x^3}{3}]_{-4}^{1}$ Thus, we have $I= [-4- \dfrac{1}{4}+16-\dfrac{7}{3}]- [ -64-64-64+\dfrac{448}{3}]=\dfrac{625}{12}$
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