Answer
a) $(r,\theta,z)=(2, \dfrac{3\pi}{4}, 1)$
b) $(r,\theta,z)=(2\sqrt 2, \dfrac{\pi}{4}, 2)$
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
$r=\sqrt{x^2+y^2} \implies r= 2$
$\tan \theta =\dfrac{y}{x} \implies \theta=\arctan (\dfrac{\sqrt 2}{-\sqrt 2})$
and $\theta=\dfrac{3\pi}{4}$
Thus, we have $(r,\theta,z)=(2, \dfrac{3\pi}{4}, 1)$
b) Here, we have
$r=\sqrt{x^2+y^2} \implies r=\sqrt{(2)^2+(2)^2}=8=2\sqrt 2$
$\tan \theta =\dfrac{y}{x} \implies \theta=\arctan (\dfrac{2}{2})$
and $\theta=\dfrac{\pi}{4}$
Thus, we have $(r,\theta,z)=(2\sqrt 2, \dfrac{\pi}{4}, 2)$
