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Most area questions are set up with vertical strips out of habit: ∫ y dx from left to right. That works until the top or bottom boundary changes halfway across the region, and then you need two integrals, or a function such as ln x that is awkward to integrate. Turning the strips sideways often fixes both. This post shows when to switch, how to set the limits, and three worked examples you can check both ways.
Where is this in the syllabus?
Section 8.2 of NCERT's Class 12 chapter on application of integrals (2026-27 reprint) gives two formulas: A = ∫ y dx for the region under y = f(x), and A = ∫ x dy for the region between x = g(y), the y-axis and the lines y = c and y = d, "considering horizontal strips". It then finds the areas of a circle and an ellipse both ways. The current edition has no separate section on the area between two curves.
The CBSE 2026-27 curriculum asks for "area under simple curves, especially lines, circles/parabolas/ellipses (in standard form only)". The JEE (Main) 2026 syllabus uses wider wording, "determining areas of the regions bounded by simple curves in standard forms", so regions bounded by a line and a parabola, as in example 2, are fair practice for JEE. Check your own year's documents.
When should you use horizontal strips?
Sketch the region, then ask one question: does the boundary on the left stay the same curve all the way up, and does the boundary on the right? If yes, horizontal strips need only one integral.
For a horizontal strip at height y:
its length is x_right − x_left, both written as functions of y;
its thickness is dy;
the limits are the lowest and highest y-values of the region.
So Area = ∫ from c to d of (x_right − x_left) dy. Right minus left plays the same role that top minus bottom plays for vertical strips.
Worked example 1: a parabola opening left
Find the area bounded by x = 4 − y² and the y-axis.
The curve meets the y-axis (x = 0) where y² = 4, so y = −2 and y = 2.
Left boundary: x = 0. Right boundary: x = 4 − y².
Area = ∫ from −2 to 2 of (4 − y²) dy = [4y − y³/3] from −2 to 2 = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 32/3.
With vertical strips you would need y = ±√(4 − x) and a square-root integral. Here it is one polynomial.
Worked example 2: a parabola and a line (JEE practice)
Find the area enclosed by y² = x and y = x − 2.
Write both as x in terms of y: x = y² and x = y + 2.
They meet where y² = y + 2, that is (y − 2)(y + 1) = 0, so y = −1 and y = 2.
At y = 0, the line gives x = 2 and the parabola x = 0, so the line is on the right.
Area = ∫ from −1 to 2 of (y + 2 − y²) dy = [y²/2 + 2y − y³/3] from −1 to 2 = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 + 7/6 = 9/2.
Now check with vertical strips. From x = 0 to 1, the region runs between y = −√x and y = √x, giving ∫ 2√x dx = 4/3. From x = 1 to 4, it runs between y = x − 2 and y = √x, giving ∫ (√x − x + 2) dx = 19/6. The total is 4/3 + 19/6 = 27/6 = 9/2. Same answer, but two integrals and an extra intersection to find.
Worked example 3: avoiding ln x
Find the area bounded by y = ln x, the y-axis and the lines y = 0 and y = 1.
Rewrite the curve as x = eʸ.
Left boundary: x = 0. Right boundary: x = eʸ. Limits: y = 0 to 1.
Area = ∫ from 0 to 1 of eʸ dy = e − 1 ≈ 1.72.
The vertical version needs a unit square from x = 0 to 1 plus ∫ from 1 to e of (1 − ln x) dx, which is e − 2. The total, 1 + (e − 2) = e − 1, agrees, but needed integration by parts.
Our post on area under curves covers vertical strips and signed area, which you need before this.
Which mistakes cost the most marks?
x-limits with dy. In example 2 the y-limits are −1 and 2, not the x-values 1 and 4.
Left minus right, which gives a negative area. Test one y-value to see which curve is on the right.
Not rewriting the curve. y = ln x must become x = eʸ before you integrate in y.
Losing the lower half. For x = 4 − y², y runs from −2, not from 0.
A practice routine
In a classroom experiment with 70 seventh-grade students, Rittle-Johnson and Star (2007) found that students who compared alternative solution methods side by side made greater gains in procedural knowledge and flexibility than students who studied the same methods one at a time. Area by both kinds of strip is a natural place to compare.
Sketch drill: for six regions, sketch and mark which strip needs fewer integrals. Do not integrate yet.
Both ways: solve three regions with both strips and confirm the answers match.
Rewrite drill: turn y = √x, y = eˣ and y = x² + 1 (for x ≥ 0) into x as a function of y.
Regions are easier to set up once you can picture the strips sweeping across them. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, find the area enclosed by x = y² and x = 2 − y² with horizontal strips, then check it with vertical strips.
