Answer
$L = \int^{4}_{1}\sqrt{1+(\frac{1}{2\sqrt{y}}-1)^{2}}dy$
Work Step by Step
$x = \sqrt{y} - y$ then $dx/dy = 1/(2\sqrt{y}) -1$ and $1+(dx/dy)^{2} = 1+(\frac{1}{2\sqrt{y}}-1)^{2}$
$L = \int^{4}_{1}\sqrt{1+(\frac{1}{2\sqrt{y}}-1)^{2}}dy$
Calculus 8th Edition
by Stewart, James
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 - Page 589: 7
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