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: 12

Answer

$$A = \frac{64}{3}$$
1517856577

Work Step by Step

Find the intersections $$x = y $$ $$4(y)+ y^{2} = 12$$ $$y^{2} + 4y -12 = 0$$ $$(y+6)(y-2)=0$$ $$y= -6.2$$ These will be the integration limits over the y-axis. The parabola will be the "upper" function; solve it for $x$. $$4x = 12 - y^{2}$$ $$x = 3-\frac{1}{4} y^{2}$$ Subtract the $x=y$ function from the parabola $$A = \displaystyle\int_{-6}^{2} \big(3-\frac{1}{4}^{2}-y\big)dy$$ $$=\left[3y-\frac{1}{12}y^{3}-\frac{1}{2}y^{2}\right]^{2}_{-6}$$ $$= 6 -\frac{8}{12}-2-(18-(18)-18)$$ $$=22-\frac{2}{3}$$ $$\frac{66-2}{3}$$ $$=\frac{64}{3}$$
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