Parameterize
by

with
and
. Take a normal vector to
,

which has norm

Then the integral of
over
is


Given that Audra was drawing a perpendicular bisector of segment AB with a compass. She started with line segment AB. Then, she placed her compass point on A and opened the compass more halfway to B.
Her next step will be "Using the width of the compass from the previous step, draw two arcs to the left and right of both A and B".
The answer is y=34
Explanation: