11.1 Polar coordinate system and transformation

Polar coordinates describe points in a plane using a distance and an angle rather than horizontal and vertical displacements. This system becomes essential in Calculus IV because many regions of integration (disks, annuli, sectors) are naturally circular, and expressing them in polar form dramatically simplifies double integrals. Here you'll build fluency with the polar system itself, coordinate transformations, and the standard polar curves you'll encounter when setting up integration bounds.

Polar Coordinate System

Defining Polar Coordinates

Every point in the plane can be located by two pieces of information: how far it is from the origin and what direction you go to reach it.

• The radius $r$ is the distance from the origin (called the pole) to the point.
• The angle $\theta$ is measured counterclockwise from the positive $x$-axis to the ray connecting the origin to the point.
• A point is written as $(r, \theta)$.

Unlike Cartesian coordinates, polar representations are not unique. Adding $2\pi$ to $\theta$ gives the same point, and the origin can be written $(0, \theta)$ for any $\theta$. Negative $r$ values are also valid: the point $(-r, \theta)$ is the same as $(r, \theta + \pi)$, meaning you go distance $r$ in the opposite direction.

Relationship to Cartesian Coordinates

The two systems are connected through trigonometry. These conversion formulas will show up constantly when you set up polar double integrals.

Polar → Cartesian:

$x = r\cos\theta$

$y = r\sin\theta$

Cartesian → Polar:

$r = \sqrt{x^2 + y^2}$

$\theta = \arctan\!\left(\frac{y}{x}\right)$

The $\theta$ formula requires care with quadrants. The function $\arctan(y/x)$ only returns values in $(-\pi/2,\, \pi/2)$, which covers quadrants I and IV. If the point lies in quadrant II or III (i.e., $x < 0$), you need to add $\pi$:

$\theta = \arctan\!\left(\frac{y}{x}\right) + \pi \quad \text{when } x < 0$

If the point is on the negative $y$-axis or at the origin, handle it by inspection. Many textbooks use $\text{atan2}(y, x)$ to sidestep this issue, but on exams you should check the quadrant manually.

Coordinate Transformation

Polar to Cartesian Transformation

1. Start with the polar point $(r, \theta)$.
2. Compute $x = r\cos\theta$.
3. Compute $y = r\sin\theta$.

Example: Convert $(3,\, \pi/4)$ to Cartesian form.

• $x = 3\cos(\pi/4) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$
• $y = 3\sin(\pi/4) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}$

The exact Cartesian coordinates are $\left(\frac{3\sqrt{2}}{2},\, \frac{3\sqrt{2}}{2}\right)$, which is approximately $(2.121,\, 2.121)$.

Cartesian to Polar Transformation

1. Start with the Cartesian point $(x, y)$.
2. Compute $r = \sqrt{x^2 + y^2}$.
3. Find the reference angle: $\alpha = \arctan\!\left(\frac{|y|}{|x|}\right)$.
4. Adjust $\theta$ based on which quadrant $(x, y)$ falls in.

Example: Convert $(-1, 1)$ to polar form.

• $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
• The reference angle is $\arctan(1/1) = \pi/4$.
• The point is in quadrant II ($x < 0,\, y > 0$), so $\theta = \pi - \pi/4 = \frac{3\pi}{4}$.
• Polar coordinates: $\left(\sqrt{2},\, \frac{3\pi}{4}\right)$.

Converting Equations (Not Just Points)

For double integrals you'll often need to convert entire equations, not just individual points. Two substitutions are especially handy:

• $x^2 + y^2 = r^2$ — turns circles centered at the origin into simple expressions in $r$.
• $x = r\cos\theta,\; y = r\sin\theta$ — handles everything else.

For instance, the circle $x^2 + y^2 = 9$ becomes $r^2 = 9$, or simply $r = 3$. The line $y = x$ becomes $r\sin\theta = r\cos\theta$, which simplifies to $\theta = \pi/4$.

Graphing in Polar Coordinates

Polar Equations

A polar equation has the form $r = f(\theta)$. For each angle $\theta$, the function tells you how far from the origin the curve extends. To sketch one by hand:

1. Build a table of $(r, \theta)$ values, sampling $\theta$ at regular intervals (every $\pi/6$ or $\pi/4$ works well).
2. Plot each point on polar graph paper (or on Cartesian axes by converting).
3. Connect the points with a smooth curve.

Graphing Techniques

Before plotting every single point, look for shortcuts:

• Symmetry about the polar axis (the positive $x$-axis): Replace $\theta$ with $-\theta$. If the equation is unchanged, the curve is symmetric about the polar axis.
• Symmetry about the line $\theta = \pi/2$: Replace $\theta$ with $\pi - \theta$. If unchanged, the curve is symmetric about the $y$-axis.
• Symmetry about the origin: Replace $r$ with $-r$. If unchanged, the curve has origin symmetry.

Using symmetry, you can often plot just half (or a quarter) of the curve and reflect.

Example: For $r = 2\cos(3\theta)$, the factor of 3 inside the cosine tells you the curve is a rose with 3 petals (odd $n$ gives $n$ petals). Each petal spans an angular width of $\pi/3$. You only need to trace $\theta \in [0, \pi]$ and the full curve emerges.

Special Polar Graphs

These curves appear frequently as boundaries in polar double integrals. Recognizing them on sight saves time when setting up integration limits.

• Circle centered at origin: $r = a$. Radius $a$, centered at the pole.
• Cardioid: $r = a(1 + \cos\theta)$ or $r = a(1 + \sin\theta)$. Heart-shaped curve that passes through the origin. The version with $\cos$ is symmetric about the $x$-axis; the version with $\sin$ is symmetric about the $y$-axis.
• Limaçon: $r = a + b\cos\theta$. When $a = b$ you get a cardioid. When $a < b$ the curve has an inner loop. When $a > b$ there's no loop, just a dimple (or none at all if $a \geq 2b$).
• Rose curves: $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$. If $n$ is odd, the curve has $n$ petals. If $n$ is even, it has $2n$ petals. Each petal has length $a$.
• Lemniscate: $r^2 = a^2\cos(2\theta)$. A figure-eight shape centered at the origin, existing only where $\cos(2\theta) \geq 0$.

Being comfortable with these curves and the transformation formulas will make setting up polar double integrals in the next sections much more straightforward.