Polar Coordinates

Polar coordinates are an alternative to rectangular coordinates for referring to points in the plane. A point in the plane has polar coordinates $(r,\theta)$ . r is (roughly) the distance from the origin to the point; $\theta$ is the angle between the radius vector for the point and the positive x-axis. (As usual, angles are positive if measured counterclockwise and negative if measured clockwise.)

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates1.eps}}$$

Example. Plot the points whose polar coordinates are: (a) $\left(2,\dfrac{3\pi}{4}\right)$ ; and (b) $\left(-2,\dfrac{7\pi}{6}\right)$ .

To plot $\left(2,\dfrac{3\pi}{4}\right)$ , rotate the x-axis through an angle of $\dfrac{3\pi}{4}$ --- counterclockwise, since the angle is positive. Then go out 2 units.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates2.eps}}$$

To plot $\left(-2,\dfrac{7\pi}{6}\right)$ , rotate the x-axis through an angle of $\dfrac{7\pi}{6}$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates3.eps}}$$

(Well, that certainly looks stupid. The numbers are upside down!) Locate the point -2 on the rotated x-axis. That is the point $\left(-2,\dfrac{7\pi}{6}\right)$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates4.eps}}$$

If you're plotting lots of points, you can do this rotation trick with a ruler, or you can make a little number line out of cardboard. If you do this, it's nearly impossible to get confused by points which have negative r's.

A point can be represented by infinitely many polar coordinate pairs.

If you add multiples of $2\pi$ to the angle, you get the same point. For example,

$$\left(17,\dfrac{51\pi}{4}\right) \quad\hbox{and}\quad \left(17,\dfrac{51\pi}{4} + 6\pi\right)$$

represent the same point in polar coordinates.

If you multiply r by -1 and add $\pi$ to $\theta$ , you get the same point.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates5.eps}}$$

Reason: Adding $\pi$ to $\theta$ and multiplying r by -1 each have the effect of "flipping" the point to the other side of the origin, so the two changes cancel out.

For example,

$$\left(17,\dfrac{51\pi}{4}\right) \quad\hbox{and}\quad \left(-17,\dfrac{51\pi}{4} + \pi\right)$$

represent the same point in polar coordinates.

This phenomenon does not occur in rectangular coordinates; it can make things like finding intersection points of polar curves a bit tricky!

You can convert between rectangular and polar using the following conversion formulas:

$$x = r\cos \theta, \quad y = r\sin \theta,$$

$$r^2 = x^2 + y^2, \quad \tan \theta = \dfrac{y}{x}.$$

All of them come from the following standard triangle:

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates6.eps}}$$

Example. Find the polar equation for the circle

$$x^2 + (y - 3)^2 = 9.$$

$$x^2 + (y - 3)^2 = 9, \quad x^2 + y^2 - 6y + 9 = 9, \quad x^2 + y^2 = 6y, \quad r^2 = 6r\sin \theta, \quad r = 6\sin \theta.\quad\halmos$$

Example. (a) Find the rectangular equation for the polar curve $r = 4
   \cos \theta$ and identify it.

Multiply by r, substitute $r^2 = x^2 + y^2$ , and complete the square:

$$r^2 = 4r \cos \theta, \quad x^2 + y^2 = 4x, \quad (x - 2)^2 + y^2 = 4.$$

This is a circle of radius 2 with center at $(2,0)$ .

(b) If you start at $\theta =
   0$ , how far do you have to go before the graph first closes up?

The entire circle is traced out as $\theta$ goes from 0 to $\pi$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates7.eps}}$$

The top half of the circle is traced out from 0 to $\dfrac{\pi}{2}$ . (Draw some radius segments from the origin out to the curve and see.) The bottom half is traced out from $\dfrac{\pi}{2}$ to $\pi$ . It winds up in the fourth quadrant instead of the second because r is negative.

Example. Find the points at which $r^2 = 2 \cos 2\theta$ and $r = 1$ intersect.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates8.eps}}$$

Solve simultaneously: $1 = 2
   \cos 2\theta$ , so $\cos 2\theta = \dfrac{1}{2}$ .

$\cos 2\theta$ equals $\dfrac{1}{2}$ when $2\theta$ is $\pm
   \dfrac{\pi}{3}$ , plus or minus multiples of $2\pi$ . I have to add or subtract multiples of $2\pi$ until $\theta$ falls outside the range $0 \le \theta \le
   2\pi$ (at which point the angles repeat).

$2\theta =
   \pm\dfrac{\pi}{3}$ gives $\theta = \pm\dfrac{\pi}{6}$ . $2\theta = 2\pi \pm
   \dfrac{\pi}{3}$ gives $\theta = \dfrac{7\pi}{6}$ and $\theta = \dfrac{5\pi}{6}$ . Adding (or subtracting) additional multiples of $2\pi$ produces values of $\theta$ outside $0
   \le \theta \le 2\pi$ , so these are the only solutions. Note that there are four intersection points in the picture.

Here's how to find $\der y
   x$ for a polar curve. The idea is to use the Chain Rule and the conversion formulas $x = r\cos \theta$ and $y = r\sin \theta$ .

$$\der y x = \dfrac{\der y \theta}{\der x \theta} = \dfrac{\der {} \theta (r\cos \theta)} {\der {} \theta (r\sin \theta)} = \dfrac{r\cos \theta + \der r \theta \sin \theta} {-r\sin \theta + \der r \theta \cos \theta}.$$

Example. Find the equation of the tangent line to $r = \sin 3\theta$ at $\theta = \dfrac{\pi}{6}$ .

When $\theta =
   \dfrac{\pi}{6}$ , $r = 1$ . The point of tangency is given by

$$x = r\cos \theta = 1\cdot \cos \dfrac{\pi}{6} = \dfrac{\sqrt{3}}{2}, \quad y = r\sin \theta = 1\cdot \sin \dfrac{\pi}{6} = \dfrac{1}{2}.$$


$$\der r \theta = 3\cos 3\theta, \quad\hbox{so at}\quad \theta = \dfrac{\pi}{6}, \quad \der r \theta = 0.$$


$$\der y x = \dfrac{(1)\left(\cos \dfrac{\pi}{6}\right) + (0)\left(\sin \dfrac{\pi}{6}\right)} {(-1)\left(\sin \dfrac{\pi}{6}\right) + (0)\left(\cos \dfrac{\pi}{6}\right)} = -\sqrt{3}.$$

The tangent line is

$$-\sqrt{3}\left(x - \dfrac{\sqrt{3}}{2}\right) = y - \dfrac{1}{2}. \quad\halmos$$

I'll give a heuristic justification for the formula for the area of the region bounded by a polar curve.

As the picture shows, a region in polar is "swept out" as if by a revolving searchlight beam.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates9.eps}}$$

Look at a small wedge-shaped piece of the region. It subtends an angle $d\theta$ and the radius is r.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates10.eps}}$$

If $d\theta$ is small, the wedge is approximately a circular wedge. The area of a circular wedge of radius r and angle $d\theta$ is $\dfrac{1}{2}r^2\,d\theta$ , so this is a good approximation to the area of the wedge-shaped piece above.

As usual, I obtain the total area by integrating to add up the areas of the little pieces:

$$A = \int_{\theta=a}^{\theta=b} \dfrac{1}{2}r^2\,d\theta.$$

Example. Here is the graph of $r = 1 - 2 \cos \theta$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates11.eps}}$$

Label the approximate points on the graph which correspond to $\theta = 0$ , $\theta =
   \dfrac{\pi}{6}$ , $\theta = \dfrac{\pi}{3}$ , $\theta = \dfrac{\pi}{2}$ , and $\theta = \pi$ . Draw arrows which indicate how the graph is traced out as $\theta$ increases. Then find the area of the region inside the outer loop but outside the inner loop.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates12.eps}}$$

Notice that the inner loop in the third quadrant is traced out as $\theta$ goes from 0 to $\dfrac{\pi}{3}$ . Why do the points wind up in the third quadrant when the angles are between 0 and $\dfrac{\pi}{3}$ ? Because for those values of $\theta$ the radius r is negative!

You can find the area inside the outer loop and outside the inner loop by subtracting the inner loop area from the outer loop area.

By symmetry, the area of the whole inner loop is twice the area of the bottom part. As we just observed, the bottom part is traced out from 0 to $\dfrac{\pi}{3}$ . So the inner loop area is

$$2 \int_0^{\pi/3} \dfrac{1}{2} (1 - 2 \cos \theta)^2\,d\theta = \int_0^{\pi/3} \left(1 - 4 \cos \theta + 4 (\cos \theta)^2\right)\,d\theta.$$

In many of these polar area problems, you'll find the double angle formulas useful in doing the integrals:

$$(\cos u)^2 = \dfrac{1}{2}(1 + \cos 2u) \quad\hbox{and}\quad (\sin u)^2 = \dfrac{1}{2}(1 - \cos 2u).$$

Continue with the inner loop area computation:

$$\int_0^{\pi/3} \left(1 - 4 \cos \theta + 4 (\cos \theta)^2\right)\,d\theta = \int_0^{\pi/3} \left(1 - 4 \cos \theta + 2 (1 + \cos 2\theta)\right)\,d\theta =$$

$$\left[\theta - 4 \sin \theta + 2\theta + \sin 2\theta\right]_0^{\pi/3} = \pi - \dfrac{3\sqrt{3}}{2}.$$

As you can see from the picture, the top outer loop is traced out from $\dfrac{\pi}{3}$ to $\pi$ . By symmetry, the area inside the whole outer loop is twice the top outer loop area.

Before I do the computation, let's clear up a common confusion. People often think that in going from $\dfrac{\pi}{3}$ to $\pi$ on the top outer loop, the inner loop stuff is somehow "automagically" subtracted.

Well, it's not so. Think about how the area formula for polar is derived. The area is divided up into thin wedges with their vertices at the origin. The area of a typical wedge is the $\dfrac{1}{2} r^2\, d\theta$ in the area formula.

Look at some wedges for the top outer loop. You can see that they extend from the origin through the inner loop.

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates13.eps}}$$

Thus, when the area formula is applied to the outer loop, it finds the area from the origin all the way out to the outer loop. It doesn't matter that the inner loop is in the way.

With that out of the way, I'll compute the outer loop area. It is

$$2 \int_{\pi/3}^\pi \dfrac{1}{2} (1 - 2 \cos \theta)^2\,d\theta = 2\pi + \dfrac{3\sqrt{3}}{2}.$$

(I'm not writing out the details, since they're essentially the same as those for the inner loop area.)

The area inside the outer loop and outside the inner loop is

$$2\pi + \dfrac{3\sqrt{3}}{2} - \left(\pi - \dfrac{3\sqrt{3}}{2}\right) = \pi + 3\sqrt{3} \approx 8.33775.\quad\halmos$$

Example. Find the area of the region inside the circle $r = \sin
   \theta$ but outside the cardioid $r = 1 - \cos \theta$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates14.eps}}$$

The region in question is the area inside the circle from 0 to $\dfrac{\pi}{2}$ minus the area inside the cardioid from 0 to $\dfrac{\pi}{2}$ .

$$\int_0^{\pi/2} \dfrac{1}{2}\left[(\sin \theta)^2 - (1 - \cos \theta)^2\right]\,d\theta = 1 - \dfrac{\pi}{4} \approx 0.21460.\quad\halmos$$

Example. Find the area of the region outside $r = \dfrac{1}{2}$ but inside $r = \cos \theta$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates15.eps}}$$

The curves intersect when $\dfrac{1}{2} = \cos \theta$ , i.e. for $\theta =
   \pm\dfrac{\pi}{3}$ . The area is

$$\int_{-\pi/3}^{\pi/3} \dfrac{1}{2}\left[(\cos \theta)^2 - \dfrac{1}{4}\right]\,d\theta = \dfrac{\sqrt{3}}{8} + \dfrac{\pi}{12} \approx 0.47831.\quad\halmos$$

Example. Find the area of the region inside $r = \cos 2\theta$ .

$$\hbox{\epsfysize=1.75in \epsffile{polar-coordinates16.eps}}$$

$$\int_0^{2\pi} \dfrac{1}{2}(\cos 2\theta)^2\,d\theta = \dfrac{1}{4}\int_0^{2\pi} \left(1 + \cos 4\theta\right)\,d\theta = \dfrac{1}{4}\left[\theta + \dfrac{1}{4}\sin 4\theta\right]_0^{2\pi} = \dfrac{\pi}{2} \approx 1.57080.\quad\halmos$$

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