Polar Graphs and Areas Revision Notes for AQA A-Level Further Maths

Polar Graphs and Areas

Introduction to polar curves

Polar curves are curves that can be defined using polar coordinates rather than the usual Cartesian (x, y) coordinates. In polar form, a point's position is determined by:

r: the distance from the origin (the pole)
θ: the angle measured anticlockwise from the positive x-axis

A polar curve is expressed as $r = f(\theta)$ , where the radius r varies as the angle θ changes. This is particularly useful for curves with circular or rotational symmetry.

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Polar coordinates provide a natural way to describe curves that exhibit symmetry around a central point. While Cartesian coordinates use (x, y) to locate a point, polar coordinates use (r, θ) - the distance and direction from the origin.

Sketching polar curves

To sketch a curve given in polar form, you need to calculate the value of r for several key values of θ. Typically, you would consider values such as $\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ .

Important curve types

Cardioid curves have the form $r = a(1 + \cos\theta)$ or $r = a(1 + \sin\theta)$ , where a is a positive constant. These curves create a distinctive heart-like shape.

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Key properties of cardioids:

The minimum value of r is 0, occurring when the trigonometric function equals -1
The maximum value of r is 2a, occurring when the trigonometric function equals 1
The curve is continuous and loops back on itself

Rose curves have the form $r = a\cos(n\theta)$ or $r = b\sin(n\theta)$ , where n is a positive integer. These curves have n loops (or petals).

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Since r represents a radius (distance from the origin), it must be positive. Therefore, when sketching rose curves, you need to identify the values of θ for which r is positive to determine where the loops appear.

Worked example 1: Sketching a cardioid

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Worked Example: Sketching a Cardioid

Problem: Sketch the curve $r = 3(1 + \cos\theta)$

Solution:

Calculate r at key angles:

When $\theta = 0$ : $r = 3(1 + 1) = 6$
When $\theta = \frac{\pi}{2}$ : $r = 3(1 + 0) = 3$
When $\theta = \pi$ : $r = 3(1 + (-1)) = 0$
When $\theta = \frac{3\pi}{2}$ : $r = 3(1 + 0) = 3$
When $\theta = 2\pi$ : $r = 3(1 + 1) = 6$

Using the identity $\cos(2\pi - x) = \cos(x)$ , we can also note:

$\cos\left(\frac{\pi}{2}\right) = \cos\left(\frac{3\pi}{2}\right) = 0$
$\cos(\pi) = -1$

This is a cardioid with:

Minimum value of r is 0 at $\theta = \pi$
Maximum value of r is 6 at $\theta = 0$ and $\theta = 2\pi$

Worked example 2: Sketching a rose curve

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Worked Example: Sketching a Rose Curve

Problem: Sketch the curve $r = 2\cos(3\theta)$

Solution:

First, find where $r = 0$ : $\cos(3\theta) = 0 \Rightarrow 3\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$

Therefore: $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

Now determine where r is positive. Consider the graph of $y = \cos(3\theta)$ :

r is positive for: $-\frac{\pi}{6} < \theta < \frac{\pi}{6}, \quad \frac{\pi}{2} < \theta < \frac{5\pi}{6}, \quad \text{and} \quad \frac{7\pi}{6} < \theta < \frac{3\pi}{2}$

The maximum value of r is 2, occurring when $\cos(3\theta) = 1$ , which happens at $\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$ .

This creates a rose curve with three loops (since n = 3). You don't need to draw all the construction half-lines in your final sketch, but you must be able to identify where they are positioned.

Area of a sector in polar coordinates

Deriving the area formula

The area of a sector of a circle with radius r and angle θ (in radians) is given by: $\text{Area} = \frac{1}{2}r^2\theta$

To find the area enclosed by a polar curve between two half-lines, we can use integration by splitting the region into many thin sectors.

Consider a polar curve $r = f(\theta)$ between the half-lines $\theta = a$ and $\theta = b$ .

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Derivation Process:

Step 1: Split the region into n thin slices, each with angle $\delta\theta$ .

Step 2: The area of each slice can be approximated as the area of a sector of a circle with radius $r = f(\theta_i)$ and angle $\delta\theta$ : $\delta A = \frac{1}{2}[f(\theta_i)]^2 \delta\theta$

Step 3: Taking the limit as $n \to \infty$ (so $\delta\theta \to 0$ ), the sum of these areas becomes an integral: $\int_a^b \frac{1}{2}[f(\theta)]^2 d\theta = \frac{1}{2}\int_a^b r^2 d\theta$

Key formula

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The area of a sector of a polar curve $r = f(\theta)$ between the half-lines $\theta = a$ and $\theta = b$ is: $A = \frac{1}{2}\int_a^b r^2 d\theta$

Critical: θ must be measured in radians for this formula to work correctly.

Worked example 3: Area of a rose curve loop

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Worked Example: Area of a Rose Curve Loop

Problem: Calculate the area enclosed by the curve $r = \sin(2\theta)$ , where $r > 0$

Solution:

First, find where $r = 0$ : $\sin(2\theta) = 0 \Rightarrow 2\theta = 0, \pi, 2\pi, 3\pi, 4\pi$ $\Rightarrow \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$

Next, determine where r is positive: $r \text{ is positive for } 0 < \theta < \frac{\pi}{2}, \quad \pi < \theta < \frac{3\pi}{2}$

Calculate the area of one loop (from $\theta = 0$ to $\theta = \frac{\pi}{2}$ ): $A = \frac{1}{2}\int_0^{\pi/2} [\sin(2\theta)]^2 d\theta$

$= \frac{1}{2}\int_0^{\pi/2} \sin^2(2\theta) d\theta$

Use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ , so $\sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2}$ :

$= \frac{1}{2}\int_0^{\pi/2} \frac{1 - \cos(4\theta)}{2} d\theta$

$= \frac{1}{4}\int_0^{\pi/2} [1 - \cos(4\theta)] d\theta$

$= \frac{1}{4}\left[\theta - \frac{\sin(4\theta)}{4}\right]_0^{\pi/2}$

$= \frac{1}{4}\left[\frac{\pi}{2} - 0 - (0 - 0)\right]$

$= \frac{\pi}{8}$

Since there are two identical loops, the total area is: $\text{Total area} = 2 \times \frac{\pi}{8} = \frac{\pi}{4}$

Area between two polar curves

To find the area between two polar curves, you need a systematic approach. This technique builds on the single-curve area formula but requires careful consideration of intersection points and curve positioning.

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Strategy for Finding Area Between Curves

Step 1: Sketch both curves and solve their equations simultaneously to find the points of intersection.

Step 2: Calculate the area of a sector for each curve using the formula $A = \frac{1}{2}\int_a^b r^2 d\theta$ , with appropriate limits.

Step 3: Add (or subtract) the areas together to find the required area, depending on the configuration.

Worked example 4: Area between two curves

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Worked Example: Area Between Two Curves

Problem: Find the area bounded by the curves $r = 3\cos\theta$ and $r = 1 + \cos\theta$

Solution:

Step 1: Find the points of intersection by solving simultaneously: $3\cos\theta = 1 + \cos\theta$ $2\cos\theta = 1$ $\cos\theta = \frac{1}{2}$ $\Rightarrow \theta = \pm\frac{\pi}{3}$

The curves intersect at $\theta = \frac{\pi}{3}$ and $\theta = -\frac{\pi}{3}$ (or equivalently $\theta = \frac{5\pi}{3}$ ).

Step 2: Calculate the area $A_1$ enclosed by $r = 1 + \cos\theta$ between $\theta = 0$ and $\theta = \frac{\pi}{3}$ :

$A_1 = \frac{1}{2}\int_0^{\pi/3} (1 + \cos\theta)^2 d\theta$

$= \frac{1}{2}\int_0^{\pi/3} [1 + 2\cos\theta + \cos^2\theta] d\theta$

Use the identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ :

$= \frac{1}{2}\int_0^{\pi/3} \left[1 + 2\cos\theta + \frac{1 + \cos(2\theta)}{2}\right] d\theta$

$= \frac{1}{2}\int_0^{\pi/3} \left[\frac{3}{2} + 2\cos\theta + \frac{\cos(2\theta)}{2}\right] d\theta$

$= \frac{1}{2}\left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin(2\theta)}{4}\right]_0^{\pi/3}$

$= \frac{1}{2}\left[\frac{3}{2} \cdot \frac{\pi}{3} + 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8} - 0\right]$

$= \frac{1}{2}\left[\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right]$

$= \frac{1}{4}\pi + \frac{9}{16}\sqrt{3}$

Step 3: Calculate the area $A_2$ enclosed by $r = 3\cos\theta$ between $\theta = \frac{\pi}{3}$ and $\theta = \frac{\pi}{2}$ :

$A_2 = \frac{1}{2}\int_{\pi/3}^{\pi/2} (3\cos\theta)^2 d\theta$

$= \frac{1}{2}\int_{\pi/3}^{\pi/2} 9\cos^2\theta d\theta$

$= \frac{9}{2}\int_{\pi/3}^{\pi/2} \frac{1 + \cos(2\theta)}{2} d\theta$

$= \frac{9}{4}\int_{\pi/3}^{\pi/2} [1 + \cos(2\theta)] d\theta$

$= \frac{9}{4}\left[\theta + \frac{\sin(2\theta)}{2}\right]_{\pi/3}^{\pi/2}$

$= \frac{9}{4}\left[\left(\frac{\pi}{2} + 0\right) - \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\right]$

$= \frac{9}{4}\left[\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right]$

$= \frac{3\pi}{8} - \frac{9\sqrt{3}}{16}$

Step 4: The total area of the top half is: $A_1 + A_2 = \frac{\pi}{4} + \frac{9\sqrt{3}}{16} + \frac{3\pi}{8} - \frac{9\sqrt{3}}{16}$

$= \frac{\pi}{4} + \frac{3\pi}{8} = \frac{5\pi}{8}$

Since the configuration is symmetrical about $\theta = 0$ , the total area is: $\text{Total area} = 2 \times \frac{5\pi}{8} = \frac{5\pi}{4}$

Common trigonometric identities

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Essential Trigonometric Identities for Polar Areas

These identities are frequently needed when calculating areas in polar coordinates:

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$
$\cos^2(x) = \frac{1 + \cos(2x)}{2}$
$\cos(2x) = 1 - 2\sin^2(x) = 2\cos^2(x) - 1$

You'll use these identities to simplify integrals involving squared trigonometric functions, which appear frequently in polar area calculations.

Exam tips

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Key Exam Tips for Success

Always use radians when working with the area formula. If angles are given in degrees, convert them first.
Sketch the curves before attempting to calculate areas. This helps you identify the correct limits and understand the geometry.
Check where r is positive for rose curves to correctly identify loop positions.
Find intersection points carefully by solving equations simultaneously when dealing with two curves.
Use symmetry to simplify calculations where possible, as seen in Example 4.
Remember the factor of ½ in the area formula - it's easy to forget under exam pressure.

Remember!

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Key Points to Remember

Polar curves are defined in the form $r = f(\theta)$ , where r is the radius and θ is the angle in radians.
Cardioids ( $r = a(1 \pm \cos\theta)$ or $r = a(1 \pm \sin\theta)$ ) have minimum value 0 and maximum value 2a.
Rose curves ( $r = a\cos(n\theta)$ or $r = b\sin(n\theta)$ ) have n loops where r is positive.
The area formula for a polar sector is $A = \frac{1}{2}\int_a^b r^2 d\theta$ , and θ must always be in radians.
When finding area between two curves, first find intersection points, then calculate separate areas and combine them appropriately.

Polar Graphs and Areas (AQA A-Level Further Maths): Revision Notes