Area of an Ellipse Using Integration: Formula, Derivation & Solved Examples

If you’ve already worked through the circle-area derivation in this chapter, the ellipse is a natural next step — the method is almost identical, with just one extra ratio to track. This post derives the area formula πab from scratch and works through several solved examples, the kind you’ll see repeatedly in CBSE Class 12 and Maharashtra Board Commerce papers.

The Equation of an Ellipse

An ellipse centered at the origin, with semi-axes a (along the x-axis) and b (along the y-axis), is written as:

x²/a² + y²/b² = 1

Like a circle, this shape is symmetric about both the x-axis and the y-axis. That symmetry is what makes the integration manageable — you only need to compute the area of one quarter and multiply by 4.

Step-by-Step Derivation

Step 1: Isolate y

Rearranging the ellipse equation for y:

y² = b²(1 − x²/a²) → y = ±(b/a)√(a² − x²)

Since we want the region in the first quadrant (above the x-axis), take the positive root:

y = (b/a)√(a² − x²)

Step 2: Set Up the First-Quadrant Integral

The first-quadrant arc of the ellipse runs from x = 0 to x = a. Using vertical strips:

Quarter Area = ∫₀ᵃ (b/a)√(a² − x²) dx

Since b/a is constant, pull it outside the integral:

Quarter Area = (b/a) ∫₀ᵃ √(a² − x²) dx

Step 3: Apply the Standard Integral

Using the same standard result from the circle derivation:

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

Step 4: Substitute the Limits

Quarter Area = (b/a) × [(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)]₀ᵃ

At x = a: (a/2)(0) + (a²/2) sin⁻¹(1) = (a²/2)(π/2) = πa²/4

At x = 0: both terms vanish.

Quarter Area = (b/a) × (πa²/4) = πab/4

Step 5: Multiply by 4 for the Full Ellipse

Total Area = 4 × (πab/4) = πab

Notice that setting a = b reduces this exactly to πa², the circle formula — a nice consistency check that confirms the derivation is correct.

Alternative Method: Horizontal Strips

Rearranging for x instead: x = (a/b)√(b² − y²)

Quarter Area = (a/b) ∫₀ᵇ √(b² − y²) dy = (a/b) × (πb²/4) = πab/4

Total Area = 4 × πab/4 = πab

Both methods converge to the same answer, as expected from the ellipse’s dual-axis symmetry.

Worked Example 1: Standard Ellipse

Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1.

Here a² = 16 so a = 4, and b² = 9 so b = 3.

Area = πab = π(4)(3) = 12π square units

Worked Example 2: Ellipse With a Larger Vertical Axis

Find the area of the region bounded by the ellipse x²/4 + y²/9 = 1.

Here a = 2 and b = 3 (note that b > a in this case, meaning the major axis lies along the y-axis rather than the x-axis).

Area = πab = π(2)(3) = 6π square units

This example is worth noting carefully: regardless of whether a or b is larger, the area formula πab stays exactly the same — only the shape’s orientation changes.

Worked Example 3: Finding the Semi-Axes From a Given Area

Sometimes a question works backward: if the area of an ellipse x²/a² + y²/9 = 1 is 15π square units, find a.

Using Area = πab with b = 3: 15π = πa(3) → a = 5

This kind of reverse problem tests whether you truly understand the formula’s structure rather than just memorizing it, so it’s worth practicing a few variations where the area or one axis is unknown instead of both being given directly.

Common Mistakes to Avoid

The most frequent error is forgetting to pull the constant ratio b/a (or a/b) out of the integral before applying the standard result — skipping this step usually leads to an answer that’s off by a factor of a or b. Another common mistake is misreading a² and b² directly as a and b without taking the square root first. Finally, students sometimes assume a must always be the “bigger” value; in reality, a and b are simply the denominators under x² and y² respectively, and either one can be larger.

Related Reading

• Area of a Circle Using Integration: Step-by-Step Derivation (πr²)

• Application of Integrals Class 12: Complete Guide to Area Under Curves

Ready to Practice?

Once you can derive πab confidently, ellipse questions become quick, reliable marks. Get AI-powered doubt resolution and topic-wise practice on ChampionsPrep for just ₹10 per use. Practice on ChampionsPrep →

Test Your Knowledge

Q1.What is the area enclosed by the ellipse x²/a² + y²/b² = 1?

πabπa²bπ(a + b)πa²/b

Q2.What is the area bounded by the ellipse x²/16 + y²/9 = 1?

7π square units24π square units12π square units25π square units

Q3.The ellipse x²/a² + y²/9 = 1 has an area of 15π square units. What is a?

351545

Q4.Setting a = b in the ellipse area formula πab gives:

2πaπaπ/a²πa²

Q5.For the ellipse x²/4 + y²/9 = 1, along which axis does the major axis lie?

The x-axisThe line y = xThe y-axisNeither axis, since a < b