JEE Mains Maths · Application of Integrals

Circles, Ellipses and Other Conics

Regions that involve a circle, an ellipse or a hyperbola, where sectors, segments and the ellipse area πab save most of the integration.

All Application of Integrals notes JEE Mains Maths strategy

Why this matters

Twenty-two PYQs, fifteen of them multiple choice, and four from 2026. Ten cut a circle with a parabola; five cut a circle with a line, a V or a second circle; seven use an ellipse, a hyperbola or another closed curve. Three ideas cover the page.

Concept 1 of 3: A circle and a parabola

Find where the circle and the parabola meet. The parabola part is a plain integral. The circle part is quicker from geometry: the area under an arc is a sector plus or minus a triangle, so

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

rarely has to be done by substitution.

Definition

$\int_0^{c}\sqrt{r^2-x^2}\,dx=\frac c2\sqrt{r^2-c^2}+\frac{r^2}2\sin^{-1}\frac cr$ .
Sector of angle $\theta$ (radians): $\frac12r^2\theta$ .
Both curves are usually symmetric about the x-axis or the y-axis: find one half and double.

Under a circular arc

\int\sqrt{r^2-x^2}\,dx=\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\frac{x}{r}+C

Worked example

Find the first-quadrant area inside

x^2+y^2=2

and above

y=x^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q62Moderate

Example 1 · Application of Integrals · Circles, Ellipses and Other Conics

The parabola

y^{2}= 4x

divides the area of the circle

x^{2}+y^{2}= 5

in two parts. The area of the smaller part is equal to :

Inside and outside a parabola

Inside

y^2=kx

means

y^2\le kx

, the side that holds the focus. Outside is the rest of the circle. Sketch both before deciding which piece to subtract from which.

Drill 9 more on a circle and a parabola

Concept 2 of 3: Segments: a circle cut by a line or a circle

A chord cuts a circle into two segments. The smaller one is a sector minus the triangle from the centre, so it needs only the angle the chord makes at the centre. The lens where two circles overlap is two segments, one from each circle, on either side of the common chord.

Definition

Chord at distance d from the centre: $\cos\frac\theta2=\frac dr$ , where $\theta$ is the angle at the centre.
Minor segment $=\frac12r^2(\theta-\sin\theta)$ ; the major segment is $\pi r^2$ minus it.
Lens between two circles = the two segments cut off by the common chord.

Minor segment

\text{segment}=\frac12r^2(\theta-\sin\theta)

Worked example

Find the area of the smaller part of

x^2+y^2=4

cut off by

x+y=2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q62Moderate

Example 2 · Application of Integrals · Circles, Ellipses and Other Conics

The area of the region enclosed between the circles

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

and

x^{2}+ (y - 2)^{2}= 4

is :

Minor or major

The formula gives the smaller segment. When the question asks for the larger portion, subtract it from

\pi r^2

— and a V-shaped cut needs the two pieces on each side added.

Drill 4 more on segments: a circle cut by a line or a circle

Concept 3 of 3: Ellipses, hyperbolas and other closed curves

An ellipse is a circle of radius a stretched by

\frac ba

in the y-direction, so its area is

\pi ab

and every region inside it scales the same way. For a hyperbola, the arc is

\sqrt{x^2-a^2}

, whose integral carries a logarithm. For any other closed curve, use its symmetry and find one quarter.

Definition

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1$ : area $\pi ab$ ; a quarter is $\frac{\pi ab}4$ .
$\int\sqrt{x^2-a^2}\,dx=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\ln\left|x+\sqrt{x^2-a^2}\right|+C$ .
Put a conic in standard form before reading off a and b.

Area of an ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1:\quad A=\pi ab

Worked example

Find the first-quadrant area inside

\frac{x^2}9+\frac{y^2}4=1

and above the chord joining

(3,0)

and

(0,2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q52Moderate

Example 3 · Application of Integrals · Circles, Ellipses and Other Conics

The area of the region, inside the ellipse

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

and outside the region bounded by the curves

y = |x| - 1

and

y = 1 - |x|

, is :

Standard form first

4x^2+9y^2=36

\frac{x^2}9+\frac{y^2}4=1

, so

a=3

b=2

and the area is

6\pi

. Reading a and b off the unscaled equation gives a wrong area.

Drill 6 more on ellipses, hyperbolas and other closed curves

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 (3)

A circle and a parabola
Under a circular arc
$\int\sqrt{r^2-x^2}\,dx=\frac{x}{2}\sqrt{r^2-x^2}+\frac{r^2}{2}\sin^{-1}\frac{x}{r}+C$
Segments: a circle cut by a line or a circle
Minor segment
$\text{segment}=\frac12r^2(\theta-\sin\theta)$
Ellipses, hyperbolas and other closed curves
Area of an ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1:\quad A=\pi ab$

Watch out for (3)

Inside and outside a parabola
→ A circle and a parabola
Minor or major
→ Segments: a circle cut by a line or a circle
Standard form first
→ Ellipses, hyperbolas and other closed curves

Test yourself on Application of Integrals

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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