Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.2 - Double Integrals over General Regions - 15.2 Exercise - Page 1061: 62

Answer

$\sin 1 -\cos 1$

Work Step by Step

We can define the domain $D$ in the Type-1 using vertical cross-sections as follows: $ D=\left\{ (x, y) | x^2 \leq y \leq 1, \ 0 \leq x \leq 1 \right\} $ Therefore, $$\iint_{D} e^{x^2} dA=\int_{0}^{1} \int_{x^2}^{1} \sqrt y \sin y \ dy \ dx $$ and we can define the domain $D$ in the Type-II using horizontal cross-sections as follows: $ D=\left\{ (x, y) | 0 \leq x \leq \sqrt y, \ 0 \leq y \leq 1 \right\} $ Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{0}^{ \sqrt y} \sqrt y \sin y \ dx \ dy \\=\int_0^1 [x\sqrt y \sin y ]_0^{\sqrt y} \ dy \\=\int_0^1 y \sin y dy$ Let us consider $a=y \implies da=dy$ and $b=-\cos y; dv =\sin y dy$ Now, $\int_0^1 y \sin y dy =\int_0^1 [-y \cos y +\int \cos y] dy \\=\int_0^1 [ -y \cos y +\sin y +C] \\=[-y \cos y +\sin y]_0^1 \\=[-1 \cos 1 +\sin 1 ]-[ -0 \cos (0) +\sin (0)] \\=\sin 1 -\cos 1$
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