EXAMPLE 1 Converting from Cylindrical to Rectangular Coordinates

Find the rectangular coordinates of the point \(P\) with cylindrical coordinates \((r,\theta,z)=\big(2,\frac{3\pi}4,5\big)\).

Solution Converting to rectangular coordinates is straightforward (Figure 12.81): \begin{align*} x &= r\cos\theta = 2\cos\frac{3\pi}4 = 2\biggl(-\frac{\sqrt 2}2\biggr)=-\sqrt 2\\ y &= r\sin\theta = 2\sin\frac{3\pi}4 = 2\biggl(\frac{\sqrt 2}2\biggr)=\sqrt 2 \end{align*}

The \(z\)-coordinate is unchanged, so \((x,y,z)= (-\sqrt 2,\sqrt 2,5)\).

EXAMPLE 2 Converting from Rectangular to Cylindrical Coordinates

Find cylindrical coordinates for the point with rectangular coordinates \((x,y,z)=(-3\sqrt{3},-3,5)\).

Solution We have \({r= \sqrt{x^2+y^2} = \sqrt{(-3\sqrt 3)^2+(-3)^2} = 6}\). The angle \(\theta\) satisfies \[ \tan\theta = \frac{y}x = \frac{-3}{-3\sqrt 3} = \frac1{\sqrt 3} \quad\Rightarrow\quad \theta = \frac{\pi}6\quad\textrm{or}\quad \frac{7\pi}6 \]

The correct choice is \(\theta = \frac{7\pi}6\) because the projection \(Q = (-3\sqrt 3,-3,0)\) lies in the third quadrant (Figure 12.82). The cylindrical coordinates are \((r,\theta,z)=\big(6,\frac{7\pi}6,5\big)\).

Note

Level Surfaces in Cylindrical Coordinates

EXAMPLE 3 Equations in Cylindrical Coordinates

Find an equation of the form \(z=f(r,\theta)\) for the surfaces

• (a) \({x^2+y^2+z^2=9}\)
• (b) \({x+y+z=1}\)

Solution We use the formulas \[ x^2+y^2=r^2,\qquad x=r\cos\theta,\qquad y=r\sin\theta \]

Figure 12.83: Level surfaces in cylindrical coordinates.

• (a) The equation \({x^2+y^2+z^2=9}\) becomes \({r^2+z^2=9}\), or \(z=\pm\sqrt{9-r^2}\). This is a sphere of radius 3.
• (b) The plane \(x+y+z=1\) becomes \[ z=1-x-y = 1-r\cos\theta-r\sin\theta\qquad \textrm{or}\qquad z = 1-r(\cos\theta+\sin\theta) \]

Note

• The symbol \(\phi\) (usually pronounced “fee,” but sometimes pronounced “fie”) is the twenty-first letter of the Greek alphabet.
• We use \(\rho\) for the radial coordinate, although \(r\) is also used to denote distance from the origin in other contexts.

Note

Spherical Coordinates \begin{align*} \rho &= \textit{distance from origin}\\ \theta &= \textit{polar angle in the \(xy\)-plane}\\ \phi &= \textit{angle of declination from the vertical} \end{align*}

In some textbooks, \(\theta\) is referred to as the azimuthal angle and \(\phi\) as the polar angle.

EXAMPLE 4 From Spherical to Rectangular Coordinates

Find the rectangular coordinates of \(P = (\rho,\theta,\phi) = \big(3,\frac{\pi}3,\frac{\pi}4\big)\), and find the radial coordinate \(r\) of its projection \(Q\) onto the \(xy\)-plane.

Solution By the formulas above, \begin{eqnarray*} x&=& \rho\cos\theta\sin\phi = 3\cos\frac{\pi}3\sin\frac{\pi}4 = 3\left(\frac12\right)\frac{\sqrt 2}2 = \frac{3\sqrt 2}4\\ y&=& \rho\sin\theta\sin\phi = 3\sin\frac{\pi}3\sin\frac{\pi}4 = 3\left(\frac{\sqrt 3}2\right)\frac{\sqrt 2}2 = \frac{3\sqrt 6}4\\ z&=& \rho \cos\phi = 3\cos\frac{\pi}4 = 3\frac{\sqrt 2}2 = \frac{3\sqrt 2}2 \end{eqnarray*}

Now consider the projection \(Q =(x,y,0)=\left(\tfrac{3\sqrt 2}4,\tfrac{3\sqrt 6}4,0\right)\) (Figure 12.86). The radial coordinate \(r\) of \(Q\) satisfies \[ r^2 = x^2+y^2 = \left(\frac{3\sqrt 2}4\right)^2 +\left(\frac{3\sqrt 6}4\right)^2 = \frac92 \]

Therefore, \(r = 3/{\sqrt 2}\).

EXAMPLE 5 From Rectangular to Spherical Coordinates

Find the spherical coordinates of the point \(P=(x,y,z)=(2,-2\sqrt3,3)\).

Solution The radial coordinate is \({\rho = \sqrt{2^2+(-2\sqrt3)^2+3^2}=\sqrt{25}=5}\). The angular coordinate \(\theta\) satisfies \[ \tan\theta =\frac{y}{x}=\frac{-2\sqrt3}2=-\sqrt3\quad \Rightarrow \quad \theta=\frac{2\pi}3 \textrm{or} \frac{5\pi}3 \]

Since the point \((x,y)= (2,-2\sqrt 3)\) lies in the fourth quadrant, the correct choice is \(\theta = \frac{5\pi}3\) (Figure 12.87). Finally, \(\cos\phi =\tfrac{z}{\rho}=\tfrac35\) and so \(\phi = \cos^{-1}\tfrac35 \approx 0.93\). Therefore, \(P\) has spherical coordinates \(\big(5,\frac{5\pi}3, 0.93\big)\).

EXAMPLE 6 Finding an Equation in Spherical Coordinates

Find an equation of the form \(\rho=f(\theta,\phi)\) for the following surfaces:

• (a) \({x^2+y^2+z^2=9}\)
• (b) \({z=x^2-y^2}\)

Solution

• (a) The equation \({x^2+y^2+z^2=9}\) defines the sphere of radius 3 centered at the origin. Since \({\rho^2=x^2+y^2+z^2}\), the equation in spherical coordinates is \(\rho=3\).
• (b) To convert \({z=x^2-y^2}\) to spherical coordinates, we substitute the formulas for \(x\), \(y\), and \(z\) in terms of \(\rho\), \(\theta\), and \(\phi\): \begin{eqnarray} \overbrace{\rho\cos\phi}^{z}&=&\overbrace{(\rho\cos\theta\sin\phi)^2}^{x^2}- \overbrace{(\rho\sin\theta\sin\phi)^2}^{y^2} \notag\\ \cos\phi&=& \rho\sin^2\phi(\cos^2\theta - \sin^2\theta)&&&(\textrm{divide by \(\rho\) and factor})\notag\\ \cos\phi &=& \rho\sin^2\phi\cos 2\theta &&&(\textrm{since \(\cos^2\theta - \sin^2\theta = \cos2\theta\)})\notag \end{eqnarray}

Solving for \(\rho\), we obtain \({ \rho = \frac{\cos\phi}{\sin^2\phi\cos2\theta}}\).

EXAMPLE 7 Spherical Coordinates via Longitude and Latitude

Find the angles \((\theta,\phi)\) for Nairobi (\(1.17^\circ\) S, \(36.48^\circ\) E) and Ottawa (\(45.27^\circ\) N, \(75.42^\circ\) W).

Solution For Nairobi, \(\theta = 36.48^\circ\) since the longitude lies to the east of Greenwich. Nairobi's latitude is south of the equator, so \(1.17 = \phi_0 - 90\) and \(\phi_0= 91.17^\circ\).

For Ottawa, we have \(\theta = 360-75.42 = 284.58^\circ\) because \(75.42^\circ\) W refers to 75.42 degrees in the negative \(\theta\) direction. Since the latitude of Ottawa is north of the equator, \(45.27 = 90- \phi_0\) and \(\phi_0 = 44.73^\circ\).