Calculus 8th Edition

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

Chapter 15 - Multiple Integrals - 15.2 Double Integrals over General Regions - 15.2 Exercises - Page 1048: 18

Answer

$\dfrac{23}{84}$

Work Step by Step

Our aim is to calculate the area of the given surface. The domain $D$ in Type-1 can be expressed as follows: $D=\left\{ (x, y) | x^3 \leq y \leq x , \ 0 \leq x \leq 1 \right\} $ Thus, $\iint_{D} (x^2+2y) \ dA=\int_{0}^{1} \int_{x^3}^{x} (x^2+2y) \ dy \ dx \\ =\int_0^1 [x^2y+y^2]_{x^3}^{x} \ dx \\= \int_0^1 [x^2 (x)+x^2-x^2 \times x^3 -(x^3)^2] d x \\ = [\dfrac{x^4}{4} ]_0^1+[\dfrac{x^3}{3} ]_0^1-[\dfrac{x^6}{6}]_0^1-[\dfrac{x^7}{7}]_0^1 \\= \dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{7} \\=\dfrac{23}{84}$
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