Answer
a) $(2, \dfrac{3\pi}{2},\dfrac{\pi}{2})$
b) $(2, \dfrac{3\pi}{4},\dfrac{3\pi}{4})$
Work Step by Step
a) $\rho=\sqrt {x^2+y^2+z^2}=\sqrt {0^2+(-2)^2+0^2}=\sqrt {0+4+0}=2$
$\cos \phi =\dfrac{0}{2}$
This gives: $\phi=\dfrac{\pi}{2}$
$\cos \theta=\dfrac{0}{2 \sin (\dfrac{\pi}{2}) }$
and $\theta=\dfrac{3\pi}{2}$
Thus, we have $(r, \theta, \phi)=(2, \dfrac{3\pi}{2},\dfrac{\pi}{2})$
b) Here, $\rho=\sqrt {(-1)^2+)(1)^2+(-\sqrt2)^2}=\sqrt {1+1+2}=2$
$\cos \phi =\dfrac{-\sqrt 2}{2}$
and $\phi=\cos^{-1}[\dfrac{-\sqrt 2}{2}]=\dfrac{3 \pi}{4}$; and $\cos \theta=\dfrac{-1}{2 \sin (\dfrac{3\pi}{4})} $
and $\theta=\cos^{-1}(\dfrac{-1}{2 \sin (\dfrac{3\pi}{4})})=\dfrac{3\pi}{4}$
Thus, we get $(r, \theta, \phi)=(2, \dfrac{3\pi}{4},\dfrac{3\pi}{4})$
