Polar Coordinates (Edexcel A-Level Further Mathematics): Revision Notes

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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 rθr\theta relationship.

Finding the Area Enclosed by a Polar Curve

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

Area=12∫αβr2 dθ\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta

Method

  1. Square rr to get r2r^2
  2. Integrate r2r^2 with respect to θ\theta
  3. Multiply the result by 1/2

Finding Tangents

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

dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}

where r′=drdθr' = \frac{dr}{d\theta}

Tangent Conditions:

  1. Parallel to the initial line (y=0y = 0)****:
dydx=0  ⟹  r′sin⁡θ+rcos⁡θ=0\frac{dy}{dx} = 0 \quad \implies r' \sin \theta + r \cos \theta = 0
  1. Perpendicular to the initial line (x=0x = 0)****:
dydx→∞  ⟹  r′cos⁡θ−rsin⁡θ=0\frac{dy}{dx} \to \infty \quad \implies r' \cos \theta - r \sin \theta = 0

Worked Examples

Note

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

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

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

Area=12∫02π32 dθ=12∫02π9 dθ\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:**

∫02π9 dθ=9θ∣02π=9(2π−0)=18π\int_0^{2\pi} 9 \, d\theta = 9\theta \Big|_0^{2\pi} = 9(2\pi - 0) = 18\pi

Step 3: Multiply by 12\frac{1}{2}:

Area=12⋅18π=9π\text{Area} = \frac{1}{2} \cdot 18\pi = 9\pi

Result: The area is 9π

Note

Example 2**: Area of a Cardioid**

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

Step 1: Set up the integral:

The cardioid completes a full cycle as θ\theta goes from 00 to 2π2\pi

Area=12∫02π(2+2cos⁡θ)2 dθ\text{Area} = \frac{1}{2} \int_0^{2\pi} (2 + 2\cos \theta)^2 \, d\theta

Step 2: Expand (2+2cos⁡θ)2(2 + 2\cos \theta)^2

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

Step 3: Substitute and simplify:

Use the identity cos⁡2θ=1+cos⁡2θ2\cos^2 \theta = \frac{1 + \cos 2\theta}{2}

4cos⁡2θ=2+2cos⁡2θ4\cos^2 \theta = 2 + 2\cos 2\theta

Substitute:

Area=12∫02π(6+8cos⁡θ+2cos⁡2θ) dθ\text{Area} = \frac{1}{2} \int_0^{2\pi} (6 + 8\cos \theta + 2\cos 2\theta) \, d\theta

Step 4: Integrate:

∫02π6 dθ=6(2π)=12π\int_0^{2\pi} 6 \, d\theta = 6(2\pi) = 12\pi∫02π8cos⁡θ dθ=0\int_0^{2\pi} 8\cos \theta \, d\theta = 0

(average value over one period)

∫02π2cos⁡2θ dθ=0\int_0^{2\pi} 2\cos 2\theta \, d\theta = 0

(average value over one period)

Total:

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

Result: The area is 6π

Note

Example 3**: Area of a Lemniscate**

Find the area enclosed by r2=9cos⁡2θr^2 = 9\cos 2\theta

Step 1: Set up the integral:

For one loop of the lemniscate, θ \theta ranges from −π4−\frac{\pi}{4} to π4\frac{\pi}{4}

Area=12∫−π4π49cos⁡2θ dθ\text{Area} = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 9\cos 2\theta \, d\theta

Step 2: Integrate:

Use ∫cos⁡2θ dθ=12sin⁡2θ\int \cos 2\theta \, d\theta = \frac{1}{2}\sin 2\theta

Area=12×9×[12sin⁡2θ]−π4π4\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 θ=π4\theta = \frac{\pi}{4}: sin⁡2θ=sin⁡π2=1\sin 2\theta = \sin \frac{\pi}{2} = 1
  • At θ=−π4\theta = -\frac{\pi}{4}: sin⁡2θ=sin⁡−π2=−1\sin 2\theta = \sin -\frac{\pi}{2} = -1 Substitute:
Area=12×9×12(1−(−1))=12×9×12×2=92\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

Note

Common Mistakes:

  1. Forgetting the 12\frac{1}{2} factor in the area formula: Always include the 1/2 in polar area calculations.

  2. Mismanaging symmetry: For curves like cardioids and lemniscates, calculate the area for one loop and use symmetry to find the total.

  3. Skipping trigonometric identities: Forgetting to simplify cos⁡2θ\cos^2 \theta or similar terms can lead to complicated integrals.

  4. Incorrect bounds: Ensure that the bounds for θ\theta correctly correspond to the desired part of the curve.

  5. Dropping negative values of rr: For r2r^2-type curves, include both +r+r and −r-r contributions to the area.

Note

Key Formulas:

  1. Area Enclosed by a Polar Curve:
Area=12∫αβr2 dθ\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta
  1. Slope of Tangent:
dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}
  1. Conditions for Tangents:
  • Parallel to the initial line (y=0y = 0)****:
r′sin⁡θ+rcos⁡θ=0r' \sin \theta + r \cos \theta = 0
  • Perpendicular to the initial line (x=0x = 0)****:
r′cos⁡θ−rsin⁡θ=0r' \cos \theta - r \sin \theta = 0
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