Answer
$r=2\sin \theta$
Work Step by Step
Here, we have $x^2+y^2-2y=0$
This implies that
$x^2+(y-1)^2=1$
Thus, we have an equation of a circle with center $(0,1)$ and radius $1$
So, $r^2-2r \sin \theta=0$
Hence, $r=2\sin \theta$
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 10 - Practice Exercises - Page 594: 34
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