JEE Mains Maths · Application of Integrals

Horizontal Strips and Sideways Parabolas

Regions whose boundaries are easier as x in terms of y, found by integrating the right curve minus the left curve along y.

All Application of Integrals notes JEE Mains Maths strategy

Why this matters

Eighteen PYQs, thirteen of them multiple choice, and two from 2026. Seven bound a sideways parabola by a line; seven are two curves, at least one sideways, where horizontal strips avoid a split or the question switches from one strip to the other; four use a tangent to a sideways parabola together with an axis. Three ideas cover the page.

Concept 1 of 3: Right minus left, integrated in y

A parabola that opens sideways, such as

y^2=4(x-1)

, is x as a function of y. Rewrite the line as x in terms of y too. Then each horizontal strip runs from the left curve to the right one, and the area is one integral in y — where vertical strips would need two.

Definition

Write each boundary as $x=g(y)$ .
The limits are the y-values where the boundaries meet.
If right minus left is $a(y-c)(d-y)$ , the area is $\frac{a(d-c)^3}{6}$ .

Horizontal strips

A=\int_{c}^{d}\big(x_{\text{right}}-x_{\text{left}}\big)\,dy

Worked example

Find the area between

y^2=x

and the line

x-y=2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q69Moderate

Example 1 · Application of Integrals · Horizontal Strips and Sideways Parabolas

The area of the region

\{(x,y):0 \leq y \leq 6 - x,\ y^{2}\geq 4x - 3,\ x \geq 0\}

is :

Limits in y, not x

Solve for the y-values where the curves meet. Substituting to get x-values first, and then using those as limits of a y-integral, gives a wrong area.

Drill 6 more on right minus left, integrated in y

Concept 2 of 3: Two sideways curves, or a choice of strips

When both curves open sideways, horizontal strips see the same left and right curve all the way, so one integral does it. More generally, if one direction of strip would switch boundaries partway, try the other direction. The area is the same either way; only the limits and the integrand change.

Definition

Two sideways curves $x=p(y)$ , $x=q(y)$ : meet where $p(y)=q(y)$ ; area $=\int(\text{right}-\text{left})\,dy$ .
If right minus left is $a(k^2-y^2)$ , the area is $\frac{4ak^3}3$ .
Choose the strip direction in which the two boundaries never change.

Symmetric pair

\int_{-k}^{k}a\,(k^2-y^2)\,dy=\frac{4ak^3}{3}

Worked example

Find the area between

y^2=x

and

y^2=4-3x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q68Moderate

Example 2 · Application of Integrals · Horizontal Strips and Sideways Parabolas

The area of the region bounded by the curves

x+ 3y^{2}= 0

and

x+ 4y^{2}= 1

is equal to :

Changing the strips changes both limits and integrand

When a question rewrites an x-integral as a y-integral, redraw the region and read each piece's left and right boundary afresh. Swapping only the limits, or only the integrand, gives a different region.

Drill 6 more on two sideways curves, or a choice of strips

Concept 3 of 3: A tangent to a sideways parabola

For

x=f(y)

, differentiate in y: the tangent at

y_0

x=f(y_0)+f'(y_0)(y-y_0)

, already in the form horizontal strips need. Because the tangent touches the parabola, the parabola minus the tangent is a perfect square, which integrates at once.

Definition

Tangent to $x=f(y)$ at $y_0$ : $x=f(y_0)+f'(y_0)(y-y_0)$ .
For $x=ay^2+by+c$ , parabola minus tangent is $a(y-y_0)^2$ .
From the x-axis up to the point of contact, the area is $\frac{a\,y_0^3}{3}$ .

Parabola minus tangent

\int_{0}^{y_0}a\,(y-y_0)^2\,dy=\frac{a\,y_0^{3}}{3}

Worked example

Find the area bounded by

y^2=4x

, its tangent at

(4,4)

and the x-axis.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q134Moderate

Example 3 · Application of Integrals · Horizontal Strips and Sideways Parabolas

A line passing through the point

A\left( - 2,0 \right)

, touches the parabola

P:y^{2}=

x- 2

at the point

B

in the first quadrant. The area, of the region bounded by the line AB , parabola P and the x -axis, is :-

dx/dy, not dy/dx

For a sideways parabola,

\frac{dx}{dy}

gives the tangent directly as x in terms of y. If you use

\frac{dy}{dx}

, its slope is the reciprocal — mixing the two gives a line that does not touch the curve.

Drill 3 more on a tangent to a sideways parabola

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)

Right minus left, integrated in y
Horizontal strips
$A=\int_{c}^{d}\big(x_{\text{right}}-x_{\text{left}}\big)\,dy$
Two sideways curves, or a choice of strips
Symmetric pair
$\int_{-k}^{k}a\,(k^2-y^2)\,dy=\frac{4ak^3}{3}$
A tangent to a sideways parabola
Parabola minus tangent
$\int_{0}^{y_0}a\,(y-y_0)^2\,dy=\frac{a\,y_0^{3}}{3}$

Watch out for (3)

Limits in y, not x
→ Right minus left, integrated in y
Changing the strips changes both limits and integrand
→ Two sideways curves, or a choice of strips
dx/dy, not dy/dx
→ A tangent to a sideways parabola

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