Answer
$r=0$
Work Step by Step
Given $$ e^{\sqrt{x^2+y^2}}=1$$
Since
$ y=r\sin\theta,\ \ x= r\cos\theta,\ \ r^2=x^2+y^2 $, then
\begin{align*}
e^{\sqrt{x^2+y^2}}&= 1\\
e^{r}&=1\\
r&=\ln 1\\
r&=0
\end{align*}
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 618: 21
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