## Calculus: Early Transcendentals 8th Edition

a) $(2, 0,\dfrac{\pi}{6})$ b) $(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$
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 $(4, \dfrac{11\pi}{6},\dfrac{\pi}{6})$