Historical Note

In 1671, Isaac Newton wrote a manuscript entitled The Method of Fluxions and Infinite Series, which contains many uses of coordinate geometry to sketch the solutions of equations. In particular, he introduces the polar coordinate system, among various other coordinate systems.

In 1691, Jacob Bernoulli published a paper also containing polar coordinates. Because Newton’s manuscript was not published until after his death in 1727, credit for the discovery of polar coordinates is usually attributed to Bernoulli.

EXAMPLE 1 From Polar to Rectangular Coordinates

Find the rectangular coordinates of point \(Q\) in Figure 11.85.

Solution The point \(Q= (r,\theta) = \big(3,\frac{5\pi}6\big)\) has rectangular coordinates: \begin{align*} x &= r\cos\theta = 3\cos\left(\frac{5\pi}6\right) =3\left(-\frac{\sqrt{3}}2\right) = -\frac{3\sqrt{3}}2\\ y &= r\sin\theta =3\sin\left(\frac{5\pi}6\right) =3\left(\frac{1}2\right) = \frac32 \end{align*}

EXAMPLE 2 From Rectangular to Polar Coordinates

Find the polar coordinates of point \(P\) in Figure 11.86.

Figure 11.86: The polar coordinates of \(P\) satisfy \(r=\sqrt{3^2+2^2}\) and \(\tan\theta=\frac23\).

Solution Since \(P=(x,y)=(3,2)\), \begin{align*} r&=\sqrt{x^2+y^2}=\sqrt{3^2+2^2}=\sqrt{13}\approx 3.6\\ \tan\theta &= \frac{y}{x} = \frac23 \end{align*} and because \(P\) lies in the first quadrant, \begin{align*} \theta &= \tan^{-1} \frac{y}x = \tan^{-1} \frac23 \approx 0.588 \end{align*} Thus, \(P\) has polar coordinates \((r,\theta) \approx (3.6,0.588)\).

Definition

By definition, \[ -\frac{\pi}2<\tan^{-1} x < \frac{\pi}2 \] If \(r>0\), a coordinate \(\theta\) of \(P=(x,y)\) is \[ \theta = \begin{cases} \tan^{-1}\frac{y}x & \text{ if \(x > 0\)} \\ \tan^{-1}\frac{y}x +\pi & \text{ if \(x<0\)}\\ \pm \frac{\pi}{2}& \text{ if \(x = 0\)} \end{cases} \]}

EXAMPLE 3 Choosing \(\theta\) Correctly

Find two polar representations of \(P = (-1,1)\), one with \(r>0\) and one with \(r<0\).

Solution The point \(P=(x,y) = (-1,1)\) has polar coordinates \((r,\theta)\), where \[ r=\sqrt{(-1)^2+1^2}=\sqrt 2,\qquad \tan\theta=\tan\frac{y}{x}=-1 \] However, \(\theta\) is not given by \[ \tan^{-1}\frac{y}x = \tan^{-1}\left(\frac1{-1}\right) = -\frac{\pi}4 \] because \(\theta = -\frac{\pi}4\) this would place \(P\) in the fourth quadrant (Figure 11.89). Since \(P\) is in the second quadrant, the correct angle is \[ \theta =\tan^{-1}\frac{y}x +\pi = -\frac{\pi}4 + \pi = \frac{3\pi}4 \] If we wish to use the negative radial coordinate \(r= -\sqrt 2\), then the angle becomes \(\theta= -\frac{\pi}4\) or \(\frac{7\pi}4\). Thus, \[ P = \left(\sqrt 2,\frac{3\pi}4\right)\qquad \textrm{or} \qquad \left(-\sqrt 2,\frac{7\pi}4\right) \]

EXAMPLE 4 Line Through the Origin

Find the polar equation of the line through the origin of slope \(\frac32\) (Figure 11.91).

Solution A line of slope \(m\) makes an angle \(\theta_0\) with the \(x\)-axis, where \(m=\tan\theta_0\). In our case, \(\theta_0 = \tan^{-1}\frac32 \approx 0.98\). The equation of the line is \(\theta= \tan^{-1} \frac32\) or \(\theta \approx 0.98\).

Figure 11.91: Line of slope \(\frac32\) through the origin.

EXAMPLE 5 Line Not Passing Through \(O\)

Show that \begin{equation} \boxed{\bbox[#fef7e5,5pt]{r=d\sec(\theta-\alpha)}}\tag{1} \end{equation} is the polar equation of the line \(L\) whose point closest to the origin is \(P_0 = (d,\alpha)\).

Figure 11.92: \(P_0\) is the point on \(L\) closest to the origin.

Solution The point \(P_0\) is obtained by dropping a perpendicular from the origin to \(L\) (Figure 11.92), and if \(P = (r,\theta)\) is any point on \(L\) other than \(P_0\), then \(\triangle \textit{OPP}_0\) is a right triangle. Therefore, \({d}/{r} = \cos(\theta - \alpha)\), or \(r = d\sec(\theta - \alpha)\), as claimed.

EXAMPLE 6

Find the polar equation of the line \(L\) tangent to the circle \(r=4\) at the point with polar coordinates \(P_0=\big(4,\frac{\pi}3\big)\).

Solution The point on \(L\) closest to the origin is \(P_0\) itself (Figure 11.93). Therefore, we take \((d,\alpha)=\big(4,\frac{\pi}3\big)\) in Eq. (1) to obtain the equation \(r=4\sec\big(\theta-\frac{\pi}3\big)\).

Figure 11.93: The tangent line has equation \(r=4\sec\left(\theta-\frac{\pi}3\right)\).

EXAMPLE 7 Converting to Rectangular Coordinates

Identify the curve with polar equation \(r=2a\cos\theta\) (\(a\) a constant).

Solution Multiply the equation by \(r\) to obtain \(r^2=2ar\cos\theta\). Because \(r^2=x^2+y^2\) and \(x=r\cos\theta\), this equation becomes \[ x^2+y^2 = 2ax\qquad\textrm{or}\qquad x^2-2ax+y^2= 0 \] Then complete the square to obtain \((x-a)^2+y^2=a^2\). This is the equation of the circle of radius \(a\) and center \((a,0)\) (Figure 11.94).

Figure 11.94

EXAMPLE 8 Symmetry About the \(x\)-Axis

Sketch the limaçon curve \newline \(r = 2\cos\theta-1\).

Solution Since \(\cos\theta\) is periodic, it suffices to plot points for \(-\pi\le \theta\le \pi\).

Step 1. Plot points

To get started, we plot points \(A\)–\(G\) on a grid and join them by a smooth curve (Figure 11.96).

Figure 11.96: Plotting \(r = 2\cos\theta-1\) using a grid.

Step 2. Analyze \(r\) as a function of \(\theta\)

For a better understanding, it is helpful to graph \(r\) as a function of \(\theta\) in rectangular coordinates. Figure 11.97 shows that

We conclude:

• The graph begins at point \(A\) in Figure 11.97(A) and moves in toward the origin as \(\theta\) varies from \(0\) to \(\frac{\pi}3\).
• Since \(r\) is negative for \(\frac{\pi}3\le\theta\le \pi\), the curve continues into the third and fourth quadrants (rather than into the first and second quadrants), moving toward the point \(G = (-3,\pi)\) in Figure 11.97(C).

Step 3. Use symmetry

Since \(r(\theta)=r(-\theta)\), the curve is symmetric with respect to the \(x\)-axis. So the part of the curve with \(-\pi\le\theta\le 0\) is obtained by reflection through the \(x\)-axis as in Figure 11.97(D).

Figure 11.97: The curve \(r = 2\cos\theta-1\) is called the limaçon, from the Latin word for “snail.” It was first described in 1525 by the German artist Albrecht Dürer.

Definition

The cylindrical coordinates \((r,\theta,z)\) of a point \((x,y,z)\) are defined by (see Figure 11.98) \begin{equation} x = r \cos \theta, y = r \sin \theta, z=z. \end{equation}

example 1

(a) Find and plot the cylindrical coordinates of (6, 6, 8). (b) If a point has cylindrical coordinates \((8,2 \pi/3, -3)\), what are its Cartesian coordinates\(?\) Plot.

solution For part (a), we have \(r = \sqrt{6^2 + 6^2} = 6 \sqrt{2}\) and \(\theta = \tan^{-1} (6/6)= \tan^{-1} (1) = \pi/4\). Thus, the cylindrical coordinates are \((6\sqrt{2}, \pi /4, 8)\). This is point P in Figure 11.100.

Figure 11.100: Some examples of the conversion between Cartesian and cylindrical coordinates.

For part (b), note that \(2\pi/3 =\pi/2 + \pi/6\) and compute \[ x = r \cos \theta = 8 \cos \frac{2 \pi}{3} = - \frac{8}{2} =-4 \] and \[ y =r \sin \theta = 8 \sin \frac{2 \pi}{3} = 8 \frac{ \sqrt{3}}{2} = 4 \sqrt{3}. \]

Thus, the Cartesian coordinates are \((-4,4 \sqrt{3},-3)\). This is point Q in the figure.

Definition

The spherical coordinates of points \((x, y, z)\) in space are the triples \((\rho, \theta, \phi)\), defined as follows: \begin{equation} x=\rho\sin \phi \cos \theta, y=\rho\sin\phi\sin\theta, z=\rho\cos \phi,\tag{3} \end{equation} where \[ \rho\geq 0, 0\,\leq\, \theta\,<\,2\pi, 0\leq \phi\leq \pi. \]

Historical Note

In 1773, Joseph Louis Lagrange was working on Newton’s gravitational theory as it applied to ellipsoids of revolution. In attempting to calculate the total gravitational attraction of such an ellipsoid, he encountered an integral that was difficult to evaluate. Motivated by this application, he introduced spherical coordinates, which allowed him to calculate the integral. We will be discussing the method of changing coordinates as it applies to multiple integrals in Section 16.2, and applications to gravitation in Section 16.3, where we show how the inverse square law of gravity allowed Newton to consider spherical masses as point masses.

Spherical coordinates are also closely connected to navigation by latitude and longitude. To see the connection, first note that the sphere of radius \(a\) centered at the origin is described by a very simple equation in spherical coordinates, namely, \(\rho = a\;\). Fixing the radius \(a\), the spherical coordinates \(\theta \) and \(\phi \) are similar to the geographic coordinates of longitude and latitude if we take the earth’s axis to be the \(z\) axis. There are differences, though: The geographical longitude is \(\vert \theta \vert \) and is called east or west longitude, according to whether \(\theta \) is a positive or negative measure from the Greenwich meridian; the geographical latitude is \(\vert \pi/2\,{-}\, \phi \vert\) and is called north or south latitude, according to whether \(\pi/2\,{-}\,\phi\) is positive or negative.

example 2

• (a) Find the spherical coordinates of the Cartesian point \((1,-1,1)\) and plot.
• (b) Find the Cartesian coordinates of the spherical coordinate point \((3, \pi /6, \pi /4)\) and plot.
• (c) Let a point have Cartesian coordinates \((2,-3, 6)\). Find its spherical coordinates and plot.
• (d) Let a point have spherical coordinates \((1, -\pi /2, \pi /4)\). Find its Cartesian coordinates and plot.

solution

• (a) \(\rho = \sqrt{x^2 + y^2 + z^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}\),
  • \(\theta = 2\pi +\tan^{-1} \!\left( \displaystyle\frac{y}{x} \right) = 2\pi +\tan^{-1}\! \left( \displaystyle\frac{-1}{1} \right) = 2\pi- \displaystyle\frac{\pi}{4}=\frac{7\pi}{4}\)
  • \(\phi= \cos^{-1} \!\left( \displaystyle\frac{z}{\rho} \right) = \cos^{-1}\! \left( \displaystyle\frac{1}{\sqrt{3}} \right) \approx 0.955 \approx 54.74^\circ.\)
  See Figure 11.103 and the formula for \(\theta\) following formula (1).
• (b) \(x = \rho \sin \phi \cos \theta = 3 \sin \!\left( \displaystyle\frac{\pi}{4} \right) \cos \!\left(\displaystyle \frac{\pi}{6} \right) =3\! \left(\displaystyle \frac{1}{\sqrt{2}} \right) \displaystyle\frac{\sqrt{3}}{2} = \displaystyle\frac{3 \sqrt{3}}{2 \sqrt{2}}\),
  • \(y = \rho \sin \phi \sin \theta = 3 \sin \! \Big(\displaystyle\frac{\pi}{4}\Big) \sin \!\left(\displaystyle\frac{\pi}{6}\right) = 3\! \left(\frac{1}{\sqrt{2}}\right)\! \left(\displaystyle\frac{1}{2} \right) = \displaystyle \frac{3}{2 \sqrt{2}}\),
  • \(z= \rho \cos \phi = 3 \cos\! \Big(\displaystyle\frac{\pi}{4}\Big) = \displaystyle\frac{3}{\sqrt{2}} = \displaystyle\frac{3\sqrt{2}}{2} .\)

    See Figure 11.103.

  57

  

  Figure 11.103: Finding (a) the spherical coordinates of the point (1, –1, 1), and (b) the Cartesian coordinates of (3, \(\pi/6\), \(\pi/4\)).
• (c) \(\rho = \sqrt{ x^2 + y^2 + z^2} = \sqrt{ 2^2 + (-3)^2 + 6^2} = \sqrt{49} = 7\),
  • \(\theta = 2\pi+\tan^{-1} \!\left(\displaystyle\frac{y}{x}\right) =2\pi+ \tan^{-1} \! \left(\displaystyle\frac{-3}{2}\right) \approx 5.3004 \hbox{ radians} \approx 303.69^\circ,\)
  • \(\phi = \cos^{-1} \! \left(\displaystyle\frac{z}{\rho} \right) = \cos^{-1}\! \left(\displaystyle\frac{6}{7} \right) \approx 0.541 \approx 31.0^\circ.\)
  See Figure 11.104.
• (d) \(x= \rho \sin \phi \cos \theta = 1 \sin\! \left(\displaystyle\frac{\pi}{4} \right) \cos \! \left(\displaystyle -\frac{\pi}{2} \right) = \bigg(\displaystyle\frac{\sqrt{2}}{2} \bigg) \,{\cdot} \ 0 =0,\)
  • \(y = \rho \sin \phi \sin \theta = 1 \sin \! \left(\displaystyle\frac{\pi}{4} \right) \sin \!\left(\displaystyle -\frac{ \pi}{2} \right)= \bigg(\displaystyle\frac{\sqrt{2}}{2} \bigg) (-1) = -\frac{\sqrt{2}}{2}\),
  • \(z = \rho \cos \phi = 1\cos \! \Big(\displaystyle\frac{\pi}{4} \Big) = \displaystyle\frac{\sqrt{2}}{2}.\)
  See Figure 11.104.
  

  Figure 11.104: Finding (a) the spherical coordinates of the point (2, \(-3\), 6), and (b) the Cartesian coordinates of (1, \(-\pi/2\), \(\pi/4\)).

example 3

Express (a) the surface \(xz=1\) and (b) the surface \(x^2 + y^2 - z^2 \,{=}\,1\) in spherical coordinates.

solution From formula (3), \(x = \rho \sin \phi \cos \theta, \hbox{and } z = \rho \cos \phi\), and so the surface \(xz = 1\) in (a) consists of all \((\rho, \theta, \phi)\) such that \[ \rho^2 \sin \phi \cos \theta \cos \phi =1, \qquad\hskip-2.5pt \hbox{that is,} \qquad \rho^2 \sin 2 \phi \cos \theta =2. \]

For part (b), we can write \[ x^2 + y^2 - z^2 = x^2 + y^2 + z^2 - 2 z^2 = \rho^2- 2 \rho^2 \cos^2 \phi, \] so that the surface is \(\rho^2 (1-2 \cos^2 \phi) =1;\) that is, \(- \rho^2 \cos\, ( 2 \phi) =1\).