Answer
$L = \int^{0}_{2} \sqrt{1+(2y-2)^{2}} dy$
Work Step by Step
$x = y^{2} - 2y$
Then $dx/ dy = 2y - 2$
and $1+(dx/dy)^2 = 1+(2y-2)^{2}$
Therefore
$L = \int^{0}_{2} \sqrt{1+(2y-2)^{2}} dy$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 6
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