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: 19

Answer

$\dfrac{11}{3}$

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) | 1 \leq y \leq 2 , \ y-1 \leq x \leq -3y+7 \right\} $ Now, $A=\iint_{D} y^2 \ dA=\int_{1}^{2} \int_{y-1}^{-3y+7} y^2 \ dx \ dy \\ =\int_{1}^{2} [xy^2]_{y-1}^{-3y+7} \ dx$ and $A= \int_1^2 (-4y^3+8y^2) \ dy \\= [\dfrac{-4y^4}{4} +\dfrac{8y^3}{3}]_1^2 \\= [-y^4+\dfrac{8y^3}{3}]_1^2\\ [-(2^4-(1)^4]-(8/3)[2^3 -1^3] \\=-15+\dfrac{56}{3} \\=\dfrac{11}{3}$
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