## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 6 - Review - Exercises: 4

#### Answer

$\frac{32}{3}$

#### Work Step by Step

Step 1 of 3: First, find the points of intersection of the curves $x=-y$ and $x=y^{2}+3y$ $y^{2}+3y=-y$ $y^{2}+4y=0$ $y \left( y+4 \right) =0$ $y=0$ or $y=-4$ The y-coordinates of the points of intersection are 0 and -4. Step 2 of 3: Sketch the curves. Find the area of the shaded region. Step 3 of 3: The area $A= \int _{-4}^{0} \left[ \left( -y \right) - \left( y^{2}+3y \right) \right] dy$ $A= \int _{-4}^{0} \left( -y-y^{2}-3y \right) dy$ $A= \int _{-4}^{0} \left( -4y-y^{2} \right) dy$ $A= \left[ -2y^{2}-\frac{y^{3}}{3} \right] _{-4}^{0}$ $A= \left[ 0-0- \left( -2 \left( -4 \right) ^{2}-\frac{ \left( -4 \right) ^{3}}{3} \right) \right]$ $A= \left[ - \left( -32+\frac{64}{3} \right) \right]$ $A=32-\frac{64}{3}$ $A=\frac{32}{3}$

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