Answer
a) $(2, 0,\dfrac{\pi}{6})$
b) $(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$
Work Step by Step
a) Convert rectangular coordinates to spherical coordinates.
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and $\rho=\sqrt {x^2+y^2+z^2}$
Here, we have $\rho=2$
Thus, $\cos \phi =\dfrac{\sqrt 3}{2} $ or, $\phi=\dfrac{\pi}{6}$;
and $\cos \theta=\dfrac{0}{2 \sin \dfrac{\pi}{6}}$
This implies that $\theta=0$
so, we get $(2, 0,\dfrac{\pi}{6})$
b) Convert rectangular coordinates to spherical coordinates.
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and $\rho=\sqrt {x^2+y^2+z^2}$
Here, we have $\rho=4$
$\phi =\cos^{-1}(\dfrac{\sqrt 3}{2})$
or, $ \phi=\dfrac{ \pi}{6}$;
and $\cos \theta=\dfrac{-1}{2 \sin \dfrac{\pi}{6}}$
or, $\theta=\dfrac{11\pi}{6}$
so, we have $(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$
