Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 13 - Multiple Integration - 13.1 Double Integrals over Rectangular Regions - 13.1 Exercises - Page 971: 20

Answer

$$\frac{\pi }{4}$$

Work Step by Step

$$\eqalign{ & \iint\limits_R {\frac{y}{{\sqrt {1 - {x^2}} }}}dA;\,\,\,\,\,\,\,R = \left\{ {\left( {x,y} \right):\frac{1}{2} \leqslant x \leqslant \frac{{\sqrt 3 }}{2},\,\,\,\,1 \leqslant y \leqslant 2} \right\} \cr & {\text{Convert to an iterated integral}} \cr & = \int_{1/2}^{\sqrt 3 /2} {\int_1^2 {\frac{y}{{\sqrt {1 - {x^2}} }}} dy} dx \cr & = \int_{1/2}^{\sqrt 3 /2} {\left[ {\int_1^2 {\frac{y}{{\sqrt {1 - {x^2}} }}} dy} \right]} dx \cr & = \int_{1/2}^{\sqrt 3 /2} {\frac{1}{{\sqrt {1 - {x^2}} }}\left[ {\int_1^2 y dy} \right]} dx \cr & {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr & \int_1^2 y dy \cr & = \left[ {\frac{{{y^2}}}{2}} \right]_1^2 \cr & {\text{evaluating the limits for the variable }}y \cr & = \frac{{{{\left( 2 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^2}}}{2} \cr & = \frac{3}{2} \cr & {\text{simplifying}} \cr & \int_{1/2}^{\sqrt 3 /2} {\frac{1}{{\sqrt {1 - {x^2}} }}\left[ {\int_1^2 y dy} \right]} dx = \int_{1/2}^{\sqrt 3 /2} {\frac{1}{{\sqrt {1 - {x^2}} }}\left( {\frac{3}{2}} \right)} dx \cr & = \frac{3}{2}\int_{1/2}^{\sqrt 3 /2} {\frac{1}{{\sqrt {1 - {x^2}} }}} dx \cr & {\text{integrating}} \cr & = \frac{3}{2}\left( {\arcsin x} \right)_{1/2}^{\sqrt 3 /2} \cr & {\text{evaluate}} \cr & = \frac{3}{2}\left( {arcsin\left( {\frac{{\sqrt 3 }}{2}} \right) - \arcsin \left( {\frac{1}{2}} \right)} \right) \cr & = \frac{3}{2}\left( {\frac{\pi }{3} - \frac{\pi }{6}} \right) \cr & = \frac{3}{2}\left( {\frac{\pi }{6}} \right) \cr & = \frac{\pi }{4} \cr} $$
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