Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 11

Answer

A = $\frac{8}{3}$

Work Step by Step

Find the area of the region enclosed by the given curves : x = $ y^{2}$-1, x = 1-$ y^2$ First find where the functions intersect: $ y^{2}$-1 = 1-$ y^2$ $2y^2$-2 = 0 y = $\sqrt ($2/2$)$ y = 1, y = -1 Then, with both graph's aspects calculated, the graphs can be drawn. By the formula, A =$\int(f(y)-g(y))dy $ we consider, f(y) = 1-$ y^2$ and g(y) = $ y^{2}$-1 (based on the graph). Then, A = $\int $( (1-$ y^2$) - ($ y^{2}$ -1 ) )dy, [-1,1] = ( y - $\frac{y^2}{3}$ - $\frac{y^2}{3}$ + y), [-1,1] = ( -$\frac{2}{3}$ + 2) - ( $\frac{2}{3}$ - 2 ) = $\frac{8}{3}$
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