Calculus: Early Transcendentals 9th Edition

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

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 10

Answer

$A = \int_{ - 1}^1 {\left( {2 - {y^2} - {y^4}} \right)} dy$

Work Step by Step

$$\eqalign{ & {\text{Let the functions }}x = {y^4}{\text{ and }}x = 2 - {y^2} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{We can find the area integrating with respect to }}y \cr & A = \int_c^d {\left[ {f\left( y \right) - g\left( y \right)} \right]} dy{\text{ }}\left( {\bf{1}} \right){\text{ }} \cr & x = 2 - {y^2} \geqslant x = {y^4}{\text{ into the interval }}\left[ { - 1,1} \right],{\text{ then}} \cr & {\text{Let }}f\left( y \right) = 2 - {y^2}{\text{ and }}g\left( y \right) = {y^4},{\text{ }}\underbrace {\left[ { - 1,1} \right]}_{\left[ {c,d} \right]} \cr & {\text{Substituting into }}\left( {\bf{1}} \right){\text{ we obtain}} \cr & A = \int_{ - 1}^1 {\left[ {\left( {2 - {y^2}} \right) - \left( {{y^4}} \right)} \right]} dy \cr & A = \int_{ - 1}^1 {\left( {2 - {y^2} - {y^4}} \right)} dy \cr} $$
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