Answer
a) $(2, 0,\dfrac{\pi}{6})$
b) $(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$
Work Step by Step
The conversion of rectangular coordinates to spherical coordinates are given as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
(a) Here, we have $\rho=2$
$\phi =\cos^{-1} [\dfrac{z}{\rho}] \implies \phi =\cos^{-1} \dfrac{\sqrt 3}{2} \implies \phi=\dfrac{\pi}{6}$; $\cos \theta=\dfrac{0}{2 \sin \dfrac{\pi}{6}} \implies \theta=0$
Thus, we have $(r, \theta, \phi)=(2, 0,\dfrac{\pi}{6})$
(b) Here,we have $\rho=4$
$\cos \phi =\dfrac{\sqrt 3}{2}$ and $ \phi=\dfrac{ \pi}{6}$
and $\cos \theta=\dfrac{-1}{2 \sin \dfrac{\pi}{6}}$ or, $ \theta=\dfrac{11\pi}{6}$
Thus, we have $(r, \theta, \phi)=(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$
