Answer
$6 \pi$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= dt$
This implies that
$L=\int_{0}^{2 \pi} (\cos t -\sin t+3) dt$
Thus, $[\sin t+\cos t+3t]_{0}^{2 \pi}=6 \pi$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 827: 22
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