Answer
a) $(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) $(4, \dfrac{2\pi}{3}, 3)$
Work Step by Step
As we know that in the cylindrical co-ordinates $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta \\ y=r \sin \theta$
a) Here, we have
$r=\sqrt{x^2+y^2} \implies r=\sqrt 2$
$x=r \cos \theta \implies -1=\sqrt 2 \cos \theta$
and $\theta=\dfrac{3\pi}{4}$
Thus, we have $(\sqrt 2, \dfrac{3\pi}{4}, 1)$
b) Here, we have
$r=\sqrt{x^2+y^2} \implies r=\sqrt{(-2)^2+(2\sqrt 3)^2}=4$
$x=r \cos \theta \implies -2=4 \cos \theta$
and $\theta=\dfrac{2\pi}{3}$
Thus, we have $(4, \dfrac{2\pi}{3}, 3)$
