10 min readConcept sheet3-mark and 5-mark questions; a favourite for the 5-mark area question

Area under circles

For x² + y² = a², use y = √(a² − x²) and the standard integral. The whole circle gives πa², four times the first-quadrant area.

Standard formx² + y² = a²: centre O(0, 0), radius a

Upper halfy = √(a² − x²), the semicircle above the x-axis

Standard integral∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + C

Values you will needsin⁻¹(1) = π/2, sin⁻¹(1/2) = π/6, sin⁻¹(√3/2) = π/3, sin⁻¹(1/√2) = π/4

The circle in standard form

The circle x² + y² = a² has its centre at the origin and radius a. It is symmetric about both axes, so the four quadrants hold equal areas. Solving for y gives y = ±√(a² − x²). The plus sign gives the upper semicircle, which is what you integrate. Only circles in this standard form are in your syllabus.

Area of the whole circle by integration

1
By symmetry, area of the circle = 4 × (area in the first quadrant).
2
First-quadrant area = ∫ y dx from 0 to a = ∫ √(a² − x²) dx from 0 to a
3
= [(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)] from 0 to a
4
At x = a: (a/2)(0) + (a²/2) sin⁻¹(1) = (a²/2)(π/2) = πa²/4
5
At x = 0: 0 + (a²/2) sin⁻¹(0) = 0
6★
So the first-quadrant area is πa²/4, and the whole area = 4 × πa²/4 = πa².

★ key step

Example 1: a quarter circle

Problem

Using integration, find the area of the region in the first quadrant bounded by the circle x² + y² = 36 and the coordinate axes.

Working

Here a = 6 and y = √(36 − x²).
A = ∫ √(36 − x²) dx from 0 to 6
= [(x/2)√(36 − x²) + 18 sin⁻¹(x/6)] from 0 to 6
= (0 + 18 × π/2) − (0 + 0)
= 9π

Answer

Area = 9π sq units.

Example 2: the smaller part cut off by a line

Problem

Find the area of the smaller part of the circle x² + y² = 4 cut off by the line x = 1.

Working

The smaller part lies between x = 1 and x = 2, equally above and below the x-axis. So A = 2 ∫ √(4 − x²) dx from 1 to 2.
Antiderivative: (x/2)√(4 − x²) + 2 sin⁻¹(x/2)
At x = 2: 0 + 2 × π/2 = π
At x = 1: (1/2)√3 + 2 × π/6 = √3/2 + π/3
So ∫ √(4 − x²) dx from 1 to 2 = π − √3/2 − π/3 = 2π/3 − √3/2
A = 2 × (2π/3 − √3/2) = 4π/3 − √3

Answer

Area = (4π/3 − √3) sq units, about 2.46 sq units.

How to draw the circle

Draw the axes and mark the centre O.
Mark (a, 0), (−a, 0), (0, a) and (0, −a), then draw the circle through them with a compass.
Draw any boundary line, such as x = 1 for x² + y² = 4, and mark where it meets the circle: (1, √3) and (1, −√3).
Shade the required region and draw one vertical strip in it.
Write the equation of the circle beside the curve.

Traps

Copy the standard integral exactly; the a²/2 in front of sin⁻¹ is often forgotten.
Use the right symmetry factor: 4 for the whole circle, 2 for a part lying equally above and below the x-axis.
Leave π and surds in the answer unless a decimal is asked for.
Sense check: no part of a circle can have more area than πa².

In real life

A round kolam or a circular well cover has area πr². Integration explains why, and it also gives the area of a part cut off by a straight edge, such as a segment of a round garden bed.

Exam angle

1 marks: “Find the value of ∫ √(9 − x²) dx from 0 to 3.” It is the area of a quarter circle of radius 3: (1/4) × π × 9 = 9π/4.

3 marks: “Using integration, find the area of the circle x² + y² = 25.” Area = 4 ∫ √(25 − x²) dx from 0 to 5 = 4[(x/2)√(25 − x²) + (25/2) sin⁻¹(x/5)] from 0 to 5 = 4 × (25/2)(π/2) = 25π sq units.

5 marks: “Using integration, find the area of the region of the circle x² + y² = 16 that lies between the lines x = −2 and x = 2.” By symmetry, A = 4 ∫ √(16 − x²) dx from 0 to 2 = 4[(x/2)√(16 − x²) + 8 sin⁻¹(x/4)] from 0 to 2 = 4[2√3 + 4π/3] = (8√3 + 16π/3) sq units.

Remember it

Half x root, plus half a-squared arcsine. That is ∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a).