Converting Between Cylindrical Coordinates and Rectangular Coordinates

1. Find the rectangular coordinates of a point \(P\) whose cylindrical coordinates are \(\left( 6,\dfrac{\pi }{3},-2\right) \).
2. Find the cylindrical coordinates of a point \(P\) whose rectangular coordinates are (\(\sqrt{3},1, 5\)).

Solution (a) We use the equations \(x=r\cos \theta \) and \(y=r\sin \theta \) with \(r=6\) and \(\theta =\dfrac{\pi }{3}.\) Then \begin{equation*} x=6\cos \dfrac{\pi }{3}=3 \qquad y=6\sin \dfrac{\pi }{3}=3\sqrt{3} \qquad z=-2 \end{equation*}

In rectangular coordinates, \(P=( 3, 3\sqrt{3},-2)\).

(b) In cylindrical coordinates \(r=\sqrt{x^2+y^2} =\sqrt{3+1}=2\). Then \(\cos\theta=\dfrac{x}{r} =\dfrac{\sqrt{3}}{2}\) and \(\sin \theta=\dfrac{y}{r}=\dfrac{1}{2}\) so \(\theta=\dfrac{\pi}{6}\). Since \(z\) remains the same, the point \(P\) in cylindrical coordinates is \((r,\theta,z) = \left(2,\dfrac{\pi}{6},5\right)\).