4 Polar Coordinates

A progressive guide to converting, graphing, differentiating, and integrating polar curves, with emphasis on symmetry, intersections, area, and arc length.

Coordinate systems and conversion

A point in the plane can be represented by $(r,\theta)$ or by $(x,y)$ . The pole is the polar analogue of the origin.

In :

$r$ is the directed distance from the pole.
$\theta$ is the directed angle measured counterclockwise from the positive $x$ -axis.
A negative value of $r$ places the point in the direction opposite the terminal side of $\theta$ .

The conversion relationships are

x=r\cos\theta,\qquad y=r\sin\theta,

and

r^2=x^2+y^2,\qquad \tan\theta=\frac{y}{x}.

When finding $\theta$ , use the signs of $x$ and $y$ to select the correct quadrant. are not unique because

(r,\theta)=(r,\theta+2\pi k)=(-r,\theta+(2k+1)\pi),

where $k$ is any integer.

Example. For the rectangular point $(-3,3\sqrt{3})$ ,

r=\sqrt{(-3)^2+(3\sqrt{3})^2}=6.

The point lies in the second quadrant, so $\theta=\frac{2\pi}{3}$ . Thus

(-3,3\sqrt{3})=\left(6,\frac{2\pi}{3}\right).

Conversely, for $(r,\theta)=\left(4,\frac{5\pi}{6}\right)$ ,

x=4\cos\frac{5\pi}{6}=-2\sqrt{3},\qquad y=4\sin\frac{5\pi}{6}=2.

Takeaway: Use $x=r\cos\theta$ and $y=r\sin\theta$ to convert to , and use $r=\sqrt{x^2+y^2}$ together with quadrant information to convert to .

Converting equations

A relates $r$ and $\theta$ . To convert from polar form to rectangular form, use

r\cos\theta=x,\qquad r\sin\theta=y,\qquad r^2=x^2+y^2.

To convert a rectangular equation to polar form, substitute

x=r\cos\theta,\\qquad y=r\sin\theta,\qquad x^2+y^2=r^2.

Important patterns include:

$r=a$ represents a circle centered at the pole with radius $|a|$ .
$\theta=\alpha$ represents a line through the pole.
$r=a\cos\theta$ becomes $x^2+y^2=ax$ .
$r=a\sin\theta$ becomes $x^2+y^2=ay$ .

Example. Starting with $r=4\sin\theta$ , multiply by $r$ :

r^2=4r\sin\theta.

Substitute $r^2=x^2+y^2$ and $r\sin\theta=y$ :

x^2+y^2=4y.

Completing the square gives

x^2+(y-2)^2=4,

which is a circle centered at $(0,2)$ with radius $2$ .

Takeaway: Multiply by $r$ when needed so that the standard substitutions for $r^2$ , $r\cos\theta$ , and $r\sin\theta$ can be applied.

Graphing polar curves

A $r=f(\theta)$ is traced by varying $\theta$ over an interval and plotting the corresponding directed radius. A reliable graphing process is:

Determine an interval that traces the curve once.
Find intercepts by solving $r=0$ and checking important angles.
Test for symmetry.
Create a table of $\theta$ - and $r$ -values.
Plot the points and connect them in the direction of increasing $\theta$ .

For $r=f(\theta)$ , useful symmetry tests are:

Replace $\theta$ by $-\theta$ to test symmetry about the polar axis.
Replace $r$ by $-r$ , or replace $\theta$ by $\theta+\pi$ , to test symmetry about the pole.
Replace $\theta$ by $\pi-\theta$ to test symmetry about the vertical line $\theta=\frac{\pi}{2}$ .

Common families include:

$r=a$ : circles centered at the pole.
$r=a\cos\theta$ and $r=a\sin\theta$ : circles through the pole.
$r=a(1\pm\cos\theta)$ and $r=a(1\pm\sin\theta)$ : cardioids.
$r=a+b\cos\theta$ and $r=a+b\sin\theta$ : limaçons.
$r=a\cos(n\theta)$ and $r=a\sin(n\theta)$ : families.
$r=a+b\theta$ : Archimedean spirals.

For the $r=a\cos(n\theta)$ or $r=a\sin(n\theta)$ , the number of petals is $n$ when $n$ is odd and $2n$ when $n$ is even. The maximum distance from the pole is $|a|$ .

Example. For

r=3\sin(2\theta),

there are four petals because $n=2$ is even. One petal is traced from $\theta=0$ to $\theta=\frac{\pi}{2}$ , where $r=0$ at both endpoints. Its maximum radius is $3$ , attained when $\theta=\frac{\pi}{4}$ .

Takeaway: Symmetry and zeros of $r$ often determine the shape and the smallest interval needed to trace a .

Slopes and tangent lines

The is found by treating the curve parametrically:

x=f(\theta)\cos\theta, \qquad y=f(\theta)\sin\theta.

Differentiating with respect to $\theta$ gives

\frac{dx}{d\theta}=\frac{dr}{d\theta}\cos\theta-r\sin\theta,

\frac{dy}{d\theta}=\frac{dr}{d\theta}\sin\theta+r\cos\theta.

Therefore,

\frac{dy}{dx}= \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta} {\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}.

A horizontal tangent requires

\frac{dy}{d\theta}=0 \quad\text{and}\quad \frac{dx}{d\theta}\ne 0.

A vertical tangent requires

\frac{dx}{d\theta}=0 \quad\text{and}\quad \frac{dy}{d\theta}\ne 0.

Both conditions must be checked. If the numerator and denominator vanish simultaneously, the point may be a cusp or singular point, or it may require additional analysis.

Example. For $r=2\sin\theta$ ,

\frac{dr}{d\theta}=2\cos\theta.

At $\theta=\frac{\pi}{4}$ , both $r$ and $\frac{dr}{d\theta}$ equal $\sqrt{2}$ . Substitution gives

\frac{dy}{d\theta}=2, \qquad \frac{dx}{d\theta}=0.

Thus the tangent is vertical.

Takeaway: Differentiate the parametric expressions for $x$ and $y$ , then identify horizontal and vertical tangents through the appropriate zero condition.

Area in

is computed by adding the areas of narrow sectors. A sector with radius $r$ and angular width $\Delta\theta$ has approximate area

\Delta A\approx\frac{1}{2}r^2\Delta\theta.

For $r=f(\theta)$ , the area traced from $\theta=\alpha$ to $\theta=\beta$ is

A=\frac{1}{2}\int_{\alpha}^{\beta}[f(\theta)]^2\,d\theta.

For the region between an outer radius and an inner radius,

A=\frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\rm out}^2-r_{\rm in}^2\right)d\theta.

The interval must trace the intended region exactly once. For a petal or another repeated component, use symmetry only after identifying the interval for one component.

Example. One petal of $r=3\sin(2\theta)$ is traced for $0\leq\theta\leq\frac{\pi}{2}$ . Its area is

A=\frac{1}{2}\int_0^{\pi/2}9\sin^2(2\theta)\,d\theta.

Using $\sin^2u=\frac{1-\cos(2u)}{2}$ , this evaluates to

A=\frac{9\pi}{8}.

Takeaway: First determine the exact tracing interval, then square the radius inside the area integral.

Arc length

is obtained by treating the parametrically. The general parametric formula is

L=\int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2}\,d\theta.

For a , simplifying the derivatives yields

\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2 =r^2+\left(\frac{dr}{d\theta}\right)^2.

Thus,

L=\int_{\alpha}^{\beta} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.

The limits must trace the intended portion exactly once.

Example. For the circle $r=a$ , $\frac{dr}{d\theta}=0$ . Over one complete revolution,

L=\int_0^{2\pi}\sqrt{a^2}\,d\theta =\int_0^{2\pi}|a|\,d\theta =2\pi|a|.

When $a>0$ , this is the familiar circumference $2\pi a$ .

Takeaway: Substitute both $r$ and $\frac{dr}{d\theta}$ into the square-root formula, and choose limits that do not retrace the curve.

Intersections and enclosed regions

A can be found systematically, but simply setting two radial expressions equal is not always sufficient because polar representations are not unique.

For $r=f(\theta)$ and $r=g(\theta)$ :

Set the radial expressions equal and solve $f(\theta)=g(\theta)$ .
Substitute each angle into either equation to find $r$ .
Check the pole separately by solving $f(\theta)=0$ and $g(\theta)=0$ .
Check equivalent representations using $(r,\theta)$ and $(-r,\theta+\pi)$ .

Example. For

r=2 \qquad\text{and}\qquad r=4\cos\theta,

equality of the radii gives

2=4\cos\theta, \qquad \cos\theta=\frac{1}{2}.

Therefore,

\theta=\frac{\pi}{3},\frac{5\pi}{3},

and the intersection points are

\left(2,\frac{\pi}{3}\right), \qquad \left(2,\frac{5\pi}{3}\right).

There is no pole intersection because $r=2$ never equals zero.

Intersections are especially useful when finding the area enclosed by two curves. On each interval between consecutive intersection angles, determine which curve is farther from the pole before applying

A=\frac{1}{2}\int \left(r_{\rm outer}^2-r_{\rm inner}^2\right)d\theta.

Takeaway: Find candidate angles, test the pole, account for equivalent polar representations, and compare radial distances before setting up an area integral.