University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 76

Answer

The area of the region is $4$.
1548952529

Work Step by Step

$x-y^2=0$ and $x+2y^2=3$ - For $x-y^2=0$, we can rewrite $$x=y^2$$ - For $x+2y^2=3$, we can rewrite $$x=3-2y^2$$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ here. 2) Find the limits of integration: To find the limits of integration, we find intersection points with respect to $y$. $$y^2=3-2y^2$$ $$3y^2=3$$ $$y^2=1$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=1$$ So the upper limit is $1$ and the lower limit is $-1$. 3) Find the area: The region from $y=-1$ to $y=1$ is bounded on the right by $x=3-2y^2$ and on the left by $x=y^2$. Therefore, the area of the region is $$A=\int^1_{-1}(3-2y^2-y^2)dy=\int^1_{-1}(3-3y^2)dy$$ $$A=(3y-y^3)\Big]^1_{-1}$$ $$A=(3-1)-(-3+1)$$ $$A=2-(-2)=4$$
Small 1548952529
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