7.1.2 Calculus with Polar Coordinates

Calculus with Polar Coordinates

Polar coordinates are useful for describing curves and calculating their properties, such as the area enclosed or the tangents to the curve. Polar calculus often uses the arc length and area formulas, tailored to the $r\theta$ relationship.

Finding the Area Enclosed by a Polar Curve

The formula for the area enclosed by a polar curve $r = f(\theta)$ between $\theta = \alpha$ and $\theta = \beta$ is:

\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta

Method

Square $r$ to get $r^2$
Integrate $r^2$ with respect to $\theta$
Multiply the result by 1/2

Finding Tangents

The slope of a tangent to a polar curve at any point can be found using:

\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

where $r' = \frac{dr}{d\theta}$

Tangent Conditions:

Parallel to the initial line ( $y = 0$ )****:

\frac{dy}{dx} = 0 \quad \implies r' \sin \theta + r \cos \theta = 0

Perpendicular to the initial line ( $x = 0$ )****:

\frac{dy}{dx} \to \infty \quad \implies r' \cos \theta - r \sin \theta = 0

Worked Examples

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Example 1: Area Enclosed by a Circle

Find the area enclosed by the circle $r = 3$

Step 1**: Set up the integral:** The curve $r = 3$ is a complete circle, so $\alpha = 0$ , $\beta = 2\pi$

\text{Area} = \frac{1}{2} \int_0^{2\pi} 3^2 \, d\theta = \frac{1}{2} \int_0^{2\pi} 9 \, d\theta

Step 2**: Integrate:**

\int_0^{2\pi} 9 \, d\theta = 9\theta \Big|_0^{2\pi} = 9(2\pi - 0) = 18\pi

Step 3: Multiply by $\frac{1}{2}$ :

\text{Area} = \frac{1}{2} \cdot 18\pi = 9\pi

Result: The area is 9π

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Example 2: Area of a Cardioid

Find the area enclosed by $r = 2 + 2\cos \theta$

Step 1: Set up the integral:

The cardioid completes a full cycle as $\theta$ goes from $0$ to $2\pi$

\text{Area} = \frac{1}{2} \int_0^{2\pi} (2 + 2\cos \theta)^2 \, d\theta

Step 2: Expand $(2 + 2\cos \theta)^2$

(2 + 2\cos \theta)^2 = 4 + 8\cos \theta + 4\cos^2 \theta

Step 3: Substitute and simplify:

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

4\cos^2 \theta = 2 + 2\cos 2\theta

Substitute:

\text{Area} = \frac{1}{2} \int_0^{2\pi} (6 + 8\cos \theta + 2\cos 2\theta) \, d\theta

Step 4: Integrate:

\int_0^{2\pi} 6 \, d\theta = 6(2\pi) = 12\pi

\int_0^{2\pi} 8\cos \theta \, d\theta = 0

(average value over one period)

\int_0^{2\pi} 2\cos 2\theta \, d\theta = 0

(average value over one period)

Total:

\text{Area} = \frac{1}{2} \times 12\pi = 6\pi

Result: The area is 6π

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Example 3: Area of a Lemniscate

Find the area enclosed by $r^2 = 9\cos 2\theta$

Step 1: Set up the integral:

For one loop of the lemniscate, $\theta$ ranges from $−\frac{\pi}{4}$ to $\frac{\pi}{4}$

\text{Area} = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 9\cos 2\theta \, d\theta

Step 2: Integrate:

Use $\int \cos 2\theta \, d\theta = \frac{1}{2}\sin 2\theta$

\text{Area} = \frac{1}{2} \times 9 \times \left[\frac{1}{2} \sin 2\theta \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}

Step 3: Evaluate bounds:

At $\theta = \frac{\pi}{4}$ : $\sin 2\theta = \sin \frac{\pi}{2} = 1$
At $\theta = -\frac{\pi}{4}$ : $\sin 2\theta = \sin -\frac{\pi}{2} = -1$ Substitute:

\text{Area} = \frac{1}{2} \times 9 \times \frac{1}{2} (1 - (-1)) = \frac{1}{2} \times 9 \times \frac{1}{2} \times 2 = \frac{9}{2}

Result: The area of one loop is 9/2

Note Summary

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Common Mistakes:

Forgetting the $\frac{1}{2}$ factor in the area formula: Always include the 1/2 in polar area calculations.
Mismanaging symmetry: For curves like cardioids and lemniscates, calculate the area for one loop and use symmetry to find the total.
Skipping trigonometric identities: Forgetting to simplify $\cos^2 \theta$ or similar terms can lead to complicated integrals.
Incorrect bounds: Ensure that the bounds for $\theta$ correctly correspond to the desired part of the curve.
Dropping negative values of $r$ : For $r^2$ -type curves, include both $+r$ and $-r$ contributions to the area.

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Key Formulas:

Area Enclosed by a Polar Curve:

\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta

Slope of Tangent:

\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

Conditions for Tangents:

Parallel to the initial line ( $y = 0$ )****:

r' \sin \theta + r \cos \theta = 0

Perpendicular to the initial line ( $x = 0$ )****:

r' \cos \theta - r \sin \theta = 0

Calculus with Polar Coordinates Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

7.1.2 Calculus with Polar Coordinates

Calculus with Polar Coordinates

Finding the Area Enclosed by a Polar Curve

Method

Finding Tangents

Tangent Conditions:

Worked Examples

Example 1: Area Enclosed by a Circle

Example 2: Area of a Cardioid

Example 3: Area of a Lemniscate

Note Summary

Common Mistakes:

Key Formulas:

500K+ Students Use These Powerful Tools to Master Calculus with Polar Coordinates For their A-Level Exams.

Other Revision Notes related to Calculus with Polar Coordinates you should explore

Calculus with Polar Coordinates Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

7.1.2 Calculus with Polar Coordinates

Calculus with Polar Coordinates

Finding the Area Enclosed by a Polar Curve

Method

Finding Tangents

Tangent Conditions:

Worked Examples

Example 1**: Area Enclosed by a Circle**

Example 2**: Area of a Cardioid**

Example 3**: Area of a Lemniscate**

Note Summary

Common Mistakes:

Key Formulas:

500K+ Students Use These Powerful Tools to Master Calculus with Polar Coordinates For their A-Level Exams.

Other Revision Notes related to Calculus with Polar Coordinates you should explore

Example 1: Area Enclosed by a Circle

Example 2: Area of a Cardioid

Example 3: Area of a Lemniscate