MHT-CET Maths · Applications of Definite Integral

Areas of Circles, Ellipses and Hyperbolas — Sectors, Segments and Standard Integrals

Circle, ellipse and hyperbola regions need one integral learnt cold — ∫√(a² − x²) dx — plus the sector and quarter-ellipse shortcuts that avoid integrating at all.

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Why this matters

9 PYQs at 44% HARD — the one page in this chapter where a formula must be memorised, because ∫√(a² − x²) dx cannot be improvised under time pressure. The recurring stems are a circle cut by a vertical line (the minor segment), the region between an ellipse's arc and its chord in the first quadrant, and a circle intersected with a parabola; the hyperbola's latus-rectum area appears once with a key that counts both branches. Half of these are answered faster by geometry — a sector, a quarter-ellipse minus a triangle — than by any integral.

Concept 1 of 4

The Circle Integral ∫√(a² − x²) dx, Quarter Discs and Sectors

Intuition

The area under

y = \sqrt{a^2 - x^2}

from

0

a

is a quarter disc,

\dfrac{\pi a^2}{4}

— which is exactly what the standard antiderivative gives at those limits. For a wedge bounded by a line through the centre, the sector formula

\dfrac12 r^2\theta

needs no integral at all.

Definition

$\int\sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$ . Learn it cold; every circle question uses it.
Quarter disc: $\int_0^a\sqrt{a^2 - x^2}\,dx = \dfrac{\pi a^2}{4}$ . So the area in the first quadrant inside $x^2 + y^2 = 4$ between $x = 0$ and $x = 2$ is $\pi$ ; the upper half of $y = \sqrt{49 - x^2}$ is $\dfrac{49\pi}{2}$ .
Sector: the region bounded by the circle $x^2 + y^2 = 4$ , the x-axis and the line $x = y\sqrt3$ (which makes $30^\circ$ with the axis) has area $\dfrac12 r^2\theta = \dfrac12\cdot4\cdot\dfrac{\pi}{6} = \dfrac{\pi}{3}$ .
Convert the line's slope to an angle: $y = \dfrac{x}{\sqrt3}$ is $\theta = \dfrac{\pi}{6}$ ; $y = x$ is $\dfrac{\pi}{4}$ .

Circle integral and sector

\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \qquad \text{sector} = \frac12 r^2\theta

Worked example

Find the area in the first quadrant bounded by the circle

x^2 + y^2 = 9

, the x-axis and the line

y = x

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Applications of Definite IntegralMODERATE

Area (in sq. units) lying in the first quadrant and bounded by the circle

x^2+y^2=4

and the lines

x=0

and

x=2

[Q114 · 10th May Shift 1 · 2023]

Reading x = y√3 as a 60° line

x = y\sqrt3

y = \dfrac{x}{\sqrt3}

, slope

\dfrac{1}{\sqrt3}

, angle

30^\circ

. The

60^\circ

reading gives

\dfrac{2\pi}{3}

, which is offered.

Drill 2 more on the circle integral ∫√(a² − x²) dx, quarter discs and sectors

Concept 2 of 4

The Smaller Segment of a Circle Cut by a Vertical Line

Intuition

The line

x = c

slices a cap off the circle. The cap is symmetric about the x-axis, so its area is twice the integral of

\sqrt{a^2 - x^2}

from

c

a

— a single application of the standard result.

Definition

Smaller part of $x^2 + y^2 = a^2$ cut off by $x = c$ ( $0 < c < a$ ): $A = 2\int_c^a\sqrt{a^2 - x^2}\,dx$ .
$c = \dfrac{a}{\sqrt2}$ : $A = 2\left[\dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}\right]_{a/\sqrt2}^{a} = 2\left[\dfrac{\pi a^2}{4} - \dfrac{a^2}{4} - \dfrac{\pi a^2}{8}\right] = \dfrac{a^2}{2}\left(\dfrac{\pi}{2} - 1\right)$ .
$x^2 + y^2 = 4$ , $x = 1$ : $2\left[\pi - \left(\dfrac{\sqrt3}{2} + \dfrac{\pi}{3}\right)\right] = \dfrac{4\pi}{3} - \sqrt3$ .
Geometric check: segment $=$ sector $-$ triangle, i.e. $\dfrac12 a^2(2\theta) - \dfrac12 a^2\sin 2\theta$ with $\cos\theta = \dfrac{c}{a}$ . For $a = 2$ , $c = 1$ : $\theta = \dfrac{\pi}{3}$ , giving $\dfrac{4\pi}{3} - 2\sin\dfrac{2\pi}{3} = \dfrac{4\pi}{3} - \sqrt3$ .

Minor segment

A = 2\int_c^a\sqrt{a^2 - x^2}\,dx = a^2\theta - \frac{a^2}{2}\sin 2\theta, \quad \cos\theta = \frac{c}{a}

Worked example

Find the area of the smaller part of the circle

x^2 + y^2 = 16

cut off by the line

x = 2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Applications of Definite IntegralMODERATE

The area of smaller part between the circle

x^{2}+y^{2}= 4

and the line

x= 1

\_\_\_\_

sq.units.

[Q115 · 26 April Shift I · 2025]

Forgetting the factor 2 for the lower half

\int_c^a\sqrt{a^2 - x^2}\,dx

is only the part above the x-axis. The segment is symmetric, so double it; the un-doubled value is always an option.

Drill 1 more on the smaller segment of a circle cut by a vertical line

Concept 3 of 4

Ellipse Arc Minus Chord: Quarter-Ellipse Minus Triangle

Intuition

In the first quadrant the ellipse

\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

and its chord

\dfrac{x}{a} + \dfrac{y}{b} = 1

both run from

(a, 0)

(0, b)

. The region between them is the quarter-ellipse minus the triangle under the chord — two known areas, no integral.

Definition

Ellipse area $= \pi ab$ ; quarter-ellipse $= \dfrac{\pi ab}{4}$ ; triangle with the axes $= \dfrac{ab}{2}$ .
Area between arc and chord $= \dfrac{\pi ab}{4} - \dfrac{ab}{2} = \dfrac{ab}{4}(\pi - 2)$ .
$a = 5$ , $b = 3$ : $\dfrac{15}{4}(\pi - 2)$ ; $a = 3$ , $b = 2$ : $\dfrac{3}{2}(\pi - 2)$ .
By integration the quarter-ellipse is $\int_0^a\dfrac{b}{a}\sqrt{a^2 - x^2}\,dx = \dfrac{b}{a}\cdot\dfrac{\pi a^2}{4}$ — the circle integral scaled by $\dfrac{b}{a}$ .

Arc minus chord

A = \frac{\pi ab}{4} - \frac{ab}{2} = \frac{ab}{4}(\pi - 2)

Worked example

Find the area between the arc of

\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1

in the first quadrant and the chord joining

(4, 0)

(0, 2)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Applications of Definite IntegralMODERATE

AOB is the positive quadrant of the ellipse

\frac{x^{2}}{25}+\frac{y^{2}}{9}= 1

in which

OA = 5,OB = 3

. The area between the arc

AB

and the chord

AB

of the ellipse in sq. units is

[Q118 · 20 April Shift II · 2025]

Using πab/2 for the quadrant

A quadrant is a QUARTER of the ellipse,

\dfrac{\pi ab}{4}

. Halving instead of quartering doubles the first term and lands on

\dfrac{15}{2}(\pi - 2)

-type distractors.

Drill 1 more on ellipse arc minus chord: quarter-ellipse minus triangle

Concept 4 of 4

Hyperbola Segments and Circle-Parabola Regions

Intuition

\int\sqrt{x^2 - a^2}\,dx

is the hyperbola's counterpart of the circle integral, with a log where the circle has an arcsine. A region bounded partly by a circle and partly by a parabola is split at the curve where the boundary changes — a half-disc plus a parabolic cap.

Definition

$\int\sqrt{x^2 - a^2}\,dx = \dfrac{x}{2}\sqrt{x^2 - a^2} - \dfrac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right|$ .
$x^2 - y^2 = 9$ : $a = b = 3$ , $e = \sqrt2$ , latus rectum at $x = 3\sqrt2$ . Area between the right branch and that chord: $2\int_3^{3\sqrt2}\sqrt{x^2 - 9}\,dx = 9\left[\sqrt2 - \log(\sqrt2 + 1)\right]$ . The official key for the 2023 sitting marks $18[\dots]$ — both branches against both latus recta; on the paper, choose the key.
$\{x^2 + y^2 \le 1\} \cap \{y^2 \le 1 - x\}$ : for $x \le 0$ the parabola condition is automatic inside the disc, giving a half-disc $\dfrac{\pi}{2}$ ; for $0 \le x \le 1$ the parabola is the tighter bound, giving $\int_0^1 2\sqrt{1 - x}\,dx = \dfrac43$ . Total $\dfrac{\pi}{2} + \dfrac43$ .
Method for any mixed region: find the x-values where the binding boundary changes, and sum simple pieces (half-disc, parabolic cap, triangle).

Hyperbola integral and the circle-parabola region

\int\sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| \qquad \{x^2 + y^2 \le 1,\ y^2 \le 1 - x\}:\ \frac{\pi}{2} + \frac43

Worked example

Find the area of the region ({(x, y): x^2 + y^2 le 4 ext{ and } y^2 le 4 - 2x}).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Applications of Definite IntegralHARD

The area (in sq. units) of the region described by

A = \{(x,y)\mid x^2+y^2 \leq 1

and

y^2 \leq 1-x\}

[Q129 · Shift 1 · 2022]

Setting up the circle-parabola region as one integral

The binding boundary is the circle for

x \le 0

and the parabola for

x \ge 0

. One integral from

-1

1

with either curve gives the wrong number; split at

x = 0

and add a half-disc to a parabolic cap.

Drill 1 more on hyperbola segments and circle-parabola regions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The Circle Integral ∫√(a² − x²) dx, Quarter Discs and Sectors
Circle integral and sector
$\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \qquad \text{sector} = \frac12 r^2\theta$
The Smaller Segment of a Circle Cut by a Vertical Line
Minor segment
$A = 2\int_c^a\sqrt{a^2 - x^2}\,dx = a^2\theta - \frac{a^2}{2}\sin 2\theta, \quad \cos\theta = \frac{c}{a}$
Ellipse Arc Minus Chord: Quarter-Ellipse Minus Triangle
Arc minus chord
$A = \frac{\pi ab}{4} - \frac{ab}{2} = \frac{ab}{4}(\pi - 2)$
Hyperbola Segments and Circle-Parabola Regions
Hyperbola integral and the circle-parabola region
$\int\sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| \qquad \{x^2 + y^2 \le 1,\ y^2 \le 1 - x\}:\ \frac{\pi}{2} + \frac43$

Watch out for (4)

Reading x = y√3 as a 60° line
→ The Circle Integral ∫√(a² − x²) dx, Quarter Discs and Sectors
Forgetting the factor 2 for the lower half
→ The Smaller Segment of a Circle Cut by a Vertical Line
Using πab/2 for the quadrant
→ Ellipse Arc Minus Chord: Quarter-Ellipse Minus Triangle
Setting up the circle-parabola region as one integral
→ Hyperbola Segments and Circle-Parabola Regions

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