## Calculus (3rd Edition)

So, $(2, \pi/3,-8) \Longrightarrow (1, \sqrt 3,-8)$.
We have $$r=2, \quad \theta =\pi/3, \quad z=-8.$$ Now, we have $$x=r\cos \theta=2 \cos \pi/3 =1,$$ $$y=r\sin \theta=2 \sin \pi/3 =\sqrt 3,$$ $$z=z=-8.$$ So, $(2, \pi/3,-8) \Longrightarrow (1, \sqrt 3,-8)$.