Answer
$( 2,\dfrac{ \pi}{6})$
Work Step by Step
The conversion of rectangular coordinates $(x,y)$ to polar coordinates $(r, \phi)$ can be computed as:
$r=\sqrt{x^2+x^2}$ and $\theta=\tan^{-1}(\dfrac{y}{x})$
We have for point $(\sqrt 3, 1)$:
$r=\sqrt{(\sqrt 3)^2+(1)^2}= \sqrt {4}=2$
and $\theta=\tan^{-1}\left(\dfrac{1}{\sqrt 3}\right)=\dfrac{ \pi}{6}$
Therefore, the point $(x,y)$ has coordinates $( 2,\dfrac{ \pi}{6})$ in the polar coordinate system.
