Area under ellipses
For x²/a² + y²/b² = 1, the upper half is y = (b/a)√(a² − x²). Four times the first-quadrant area gives the whole ellipse, πab.
The ellipse in standard form
An ellipse x²/a² + y²/b² = 1 looks like a circle stretched along one axis. It meets the x-axis at (±a, 0) and the y-axis at (0, ±b). If a > b it is longer along the x-axis; if b > a it is longer along the y-axis. Always change the equation to this form first: for example, 4x² + 9y² = 36 becomes x²/9 + y²/4 = 1.
Area of the whole ellipse by integration
- 1From x²/a² + y²/b² = 1, y² = (b²/a²)(a² − x²), so the upper half is y = (b/a)√(a² − x²).
- 2Area = 4 × first-quadrant area = 4 ∫ (b/a)√(a² − x²) dx from 0 to a
- 3= (4b/a)[(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)] from 0 to a
- 4= (4b/a)(a²/2)(π/2)
- 5★= πab
★ key step
Example 1: an ellipse longer along the y-axis
Using integration, find the area enclosed by the ellipse 25x² + 4y² = 100.
- Divide by 100: x²/4 + y²/25 = 1. So a = 2 and b = 5.
- Upper half: y = (5/2)√(4 − x²).
- Area = 4 ∫ (5/2)√(4 − x²) dx from 0 to 2 = 10 ∫ √(4 − x²) dx from 0 to 2
- = 10[(x/2)√(4 − x²) + 2 sin⁻¹(x/2)] from 0 to 2
- = 10 × (2 × π/2) = 10π
Area = 10π sq units. Check: πab = π × 2 × 5 = 10π.
Example 2: a band between two lines
Find the area of the region of the ellipse x²/36 + y²/9 = 1 that lies between the lines x = −3 and x = 3.
- a = 6 and b = 3, so the upper half is y = (3/6)√(36 − x²) = (1/2)√(36 − x²).
- By symmetry about both axes, A = 4 ∫ (1/2)√(36 − x²) dx from 0 to 3 = 2 ∫ √(36 − x²) dx from 0 to 3.
- = 2[(x/2)√(36 − x²) + 18 sin⁻¹(x/6)] from 0 to 3
- At x = 3: (3/2)√27 + 18 × π/6 = (3/2)(3√3) + 3π = 9√3/2 + 3π
- At x = 0: 0
- A = 2(9√3/2 + 3π) = 9√3 + 6π
Area = (9√3 + 6π) sq units, about 34.4 sq units.
Circle vs ellipse
The ellipse integral is the circle integral multiplied by b/a. Put b = a and you get the circle back.
How to draw the ellipse
- Write the equation in standard form and read off a and b.
- Mark (a, 0), (−a, 0), (0, b) and (0, −b) on the axes.
- Join them with a smooth oval, longest along the axis with the larger of a and b.
- Draw any boundary lines, shade the region, and draw one vertical strip.
- Write the equation beside the curve.
The outline of many cricket grounds and running tracks is close to an ellipse. A groundsman can estimate the turf needed with πab before ordering it.
1 marks: “Write the area of the ellipse x²/49 + y²/4 = 1.” a = 7 and b = 2, so the area is πab = 14π sq units.
3 marks: “Using integration, find the area of the region bounded by the ellipse x²/9 + y²/4 = 1 in the first quadrant.” y = (2/3)√(9 − x²). A = (2/3) ∫ √(9 − x²) dx from 0 to 3 = (2/3)[(x/2)√(9 − x²) + (9/2) sin⁻¹(x/3)] from 0 to 3 = (2/3)(9π/4) = 3π/2 sq units.
5 marks: “Using integration, find the area enclosed by the ellipse 16x² + 25y² = 400.” x²/25 + y²/16 = 1, so a = 5, b = 4 and y = (4/5)√(25 − x²). A = 4 × (4/5) ∫ √(25 − x²) dx from 0 to 5 = (16/5)(25π/4) = 20π sq units.
Circle πa², ellipse πab. Same integral, just multiply by b/a.