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: 84

Answer

The area of the region is $8$.
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Work Step by Step

$x+y^2=3$ and $4x+y^2=0$ - For the curve $x+y^2=3$, we can rewrite into $$x=-y^2+3$$ - For the curve $4x+y^2=0$, we can rewrite into $$x=-\frac{y^2}{4}$$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ in this exercise. 2) Find limits of integration: To find the limits of integration, we will find the intersection points with respect to $y$. $$-y^2+3=-\frac{y^2}{4}$$ $$4y^2-12=y^2$$ $$3y^2=12$$ $$y^2=4$$ $$y=-2\hspace{1cm}\text{or}\hspace{1cm}y=2$$ Therefore, the upper limit is $2$ and the lower one is $-2$. 3) Find the area: The region from $y=-2$ to $y=2$ is bounded on the right by $x=-y^2+3$ and on the left by $x=-y^2/4$. So according to definition, the area of the region is $$A=\int^2_{-2}(-y^2+3+\frac{y^2}{4})dy=\int^2_{-2}(-\frac{3y^2}{4}+3)dy$$ $$A=-\frac{y^3}{4}+3y\Big]^2_{-2}$$ $$A=(-2+6)-(2-6)$$ $$A=2(-2+6)=8$$
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