Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 22

Answer

$\frac{1}{27} (80\sqrt{10} - 13\sqrt{13})$

Work Step by Step

$x^{2} = (y-4)^{3}, x = (y-4)^{3/2} $(for $x>0$) then $x' = \frac{3}{2} (y-4)^{1/2}$ and $1+(dx/dy)^{2} = 1+ \frac{9}{4} (y-4) = \frac{9}{4}y - 8$ So $L = \int^{8}_{5} \sqrt{\frac{9}{4}y-8} dy = \int^{10}_{13/4} \sqrt{u} (\frac{4}{9} du) = \frac{4}{9}[\frac{2}{3} u^{3/2}]^{10}_{13/4} = \frac{1}{27} (80\sqrt{10} - 13\sqrt{13})$
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