Answer
a) $(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) $(4, \dfrac{2\pi}{3}, 3)$
Work Step by Step
In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$
Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
a) $r=\sqrt{x^2+y^2} \implies r=\sqrt 2$
$\tan \theta=\dfrac{y}{x}$
$\implies \theta=\arctan (\dfrac{-1}{2})=\dfrac{3\pi}{4}$
Hence, $(r,\theta,z)=(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) $ r=\sqrt{(-2)^2+(2\sqrt 3)^2}=\sqrt{4+12}=4$
$x=r \cos \theta \implies -2=4 \cos \theta$
This gives: $\theta=\dfrac{2\pi}{3}$
Hence, $(r,\theta,z)=(4, \dfrac{2\pi}{3}, 3)$
