Areas of Circles and Ellipses

The Circle: $\pi a^2$ Earned by Integration

The formula "area of a circle $= \pi r^2$" is stated without proof in earlier classes. This chapter derives it — the first big payoff of the strip method.

Setting up with symmetry

The circle $x^2 + y^2 = a^2$ is symmetric about both axes, so

$$\text{whole area} = 4 \times (\text{area of the first-quadrant region AOBA})$$

where AOBA is bounded by the curve, the $x$-axis and the ordinates $x = 0$, $x = a$.

Vertical strips

Solving $x^2 + y^2 = a^2$ gives $y = \pm\sqrt{a^2 - x^2}$; in the first quadrant $y$ is positive, so the strip height is $y = \sqrt{a^2 - x^2}$:

$$A = 4\int_0^a \sqrt{a^2 - x^2}\,dx = 4\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$$

At $x = a$ the first term vanishes and $\sin^{-1}1 = \frac{\pi}{2}$; at $x = 0$ both terms vanish:

$$A = 4\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi a^2$$

Horizontal strips — the same answer

Slicing the same quadrant into horizontal strips of length $x = \sqrt{a^2 - y^2}$:

$$A = 4\int_0^a x\,dy = 4\int_0^a\sqrt{a^2 - y^2}\,dy = \pi a^2$$

The integral is identical with $y$ as the variable — a live demonstration that the strip direction is a choice, not a constraint.

Key Point: The engine inside both computations is the Chapter 7 root formula $\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$. One quadrant contributes $\frac{\pi a^2}{4}$ — worth remembering on its own: $\int_0^a\sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$, a quarter circle.

[JEE Tip] Never re-derive $\int_0^a\sqrt{a^2 - x^2}\,dx$ in the exam — read it geometrically as a quarter-circle area $\frac{\pi a^2}{4}$. Fragments like $\int_0^2\sqrt{4 - x^2}\,dx = \pi$ then take five seconds.

The Ellipse: $\pi ab$

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ the same symmetry argument applies — the curve is symmetric about both axes, so the whole area is $4\times$ the first-quadrant region.

Vertical strips

Solving for $y$ (positive root in the first quadrant): $y = \frac{b}{a}\sqrt{a^2 - x^2}$. So

$$A = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\,dx = \frac{4b}{a}\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a = \frac{4b}{a}\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi ab$$

Horizontal strips

With $x = \frac{a}{b}\sqrt{b^2 - y^2}$ and $y$ running $0$ to $b$:

$$A = 4\int_0^b \frac{a}{b}\sqrt{b^2 - y^2}\,dy = \frac{4a}{b}\cdot\frac{\pi b^2}{4} = \pi ab$$

Reading the result

1. The ellipse is a circle stretched by $\frac{b}{a}$ vertically — every strip height scales by $\frac ba$, so the area scales the same way: $\pi a^2 \cdot \frac{b}{a} = \pi ab$.
2. Setting $b = a$ recovers the circle: $\pi a\cdot a = \pi a^2$ — a built-in consistency check.
3. For a numerical ellipse like $\frac{x^2}{16} + \frac{y^2}{9} = 1$: read off $a = 4$, $b = 3$ (the square roots of the denominators), giving area $12\pi$.

Key Point: For any standard-form ellipse, area $= \pi ab$ where $a^2$ and $b^2$ are the denominators. The only working error available is forgetting the square roots — $\frac{x^2}{16} + \frac{y^2}{9} = 1$ has area $\pi\cdot4\cdot3 = 12\pi$, never $\pi\cdot16\cdot9$.

[JEE Tip] Exam variants ask for parts of these regions: a quadrant of the ellipse ($\frac{\pi ab}{4}$), the region between the ellipse and a chord, or the circle piece cut by a vertical line. The strip setup is unchanged — only the limits move. Sketch, mark the limits, and reuse the same root integral.

Solved Examples

Example 1: The circle, in full

Find the area enclosed by the circle $x^2 + y^2 = a^2$.

Solution:

1. Symmetry: whole area $= 4\times$ area of the first-quadrant region AOBA (curve, $x$-axis, $x = 0$ to $x = a$).
2. Strip: $y = \sqrt{a^2 - x^2}$ (positive root in the first quadrant): $A = 4\int_0^a\sqrt{a^2 - x^2}\,dx$.
3. Root formula: $4\left[\frac x2\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac xa\right]_0^a = 4\left(0 + \frac{a^2}{2}\cdot\frac{\pi}{2}\right)$.

Final Answer: $\pi a^2$ square units.

Example 2: The circle by horizontal strips

Repeat Example 1 using horizontal strips.

Solution:

1. Strip: at height $y$, the strip runs from the $y$-axis to the curve: length $x = \sqrt{a^2 - y^2}$.
2. Set up: $A = 4\int_0^a x\,dy = 4\int_0^a\sqrt{a^2 - y^2}\,dy$.
3. Evaluate: the same integral as before with $y$ as the variable: $4\cdot\frac{\pi a^2}{4} = \pi a^2$.

Final Answer: $\pi a^2$ square units — identical, as it must be.

Takeaway: When both directions work, the integrals are often literally the same. The choice matters only when the curve solves cleanly for one variable but not the other.

Example 3: The ellipse

Find the area enclosed by the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Solution:

1. Symmetry: $A = 4\times$ first-quadrant area.
2. Strip: $y = \frac ba\sqrt{a^2 - x^2}$: $A = \frac{4b}{a}\int_0^a\sqrt{a^2 - x^2}\,dx$.
3. Quarter-circle value: $\int_0^a\sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$, so $A = \frac{4b}{a}\cdot\frac{\pi a^2}{4}$.

Final Answer: $\pi ab$ square units.

Example 4: A numerical ellipse

Find the area of the region bounded by the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$.

Solution:

1. Identify $a$ and $b$: $a^2 = 16 \Rightarrow a = 4$; $b^2 = 9 \Rightarrow b = 3$.
2. Apply the formula: $A = \pi ab = \pi\cdot4\cdot3$.

Final Answer: $12\pi$ square units.

Example 5: A taller-than-wide ellipse

Find the area of the region bounded by the ellipse $\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$.

Solution:

1. Identify: $a = 2$ (along $x$), $b = 3$ (along $y$) — here the major axis is vertical, which changes nothing in the formula.
2. Apply: $A = \pi\cdot2\cdot3$.

Final Answer: $6\pi$ square units.

Takeaway: $\pi ab$ is orientation-blind — it doesn't care which of $a, b$ is larger.

Example 6: A semicircular region

Find the area bounded by the curve $y = \sqrt{16 - x^2}$ and the $x$-axis.

Solution:

1. Recognise the curve: $y = \sqrt{16 - x^2}$ is the upper half of the circle $x^2 + y^2 = 16$ (radius $4$), meeting the $x$-axis at $x = \pm4$.
2. Set up: $A = \int_{-4}^{4}\sqrt{16 - x^2}\,dx = 2\int_0^4\sqrt{16 - x^2}\,dx$ (even integrand).
3. Quarter-circle value: $\int_0^4\sqrt{16 - x^2}\,dx = \frac{\pi\cdot16}{4} = 4\pi$, so $A = 8\pi$.
4. Check: semicircle of radius $4$: $\frac12\pi\cdot16 = 8\pi$. ✓

Final Answer: $8\pi$ square units.

Example 7: A quadrant with explicit working

Find the area of the region in the first quadrant enclosed by $x^2 + y^2 = 1$, the $x$-axis and the $y$-axis.

Solution:

1. Set up: $A = \int_0^1\sqrt{1 - x^2}\,dx$.
2. Root formula: $\left[\frac x2\sqrt{1 - x^2} + \frac12\sin^{-1}x\right]_0^1 = 0 + \frac12\cdot\frac{\pi}{2}$.
3. Conclude: $\frac{\pi}{4}$ — a quarter of the unit circle's area $\pi$. ✓

Final Answer: $\dfrac{\pi}{4}$ square units.

Takeaway: Fragments of circles are everywhere in this chapter's exercises. Train the reflex: $\int_0^a\sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4}$ (quarter circle), $\int_{-a}^{a} = \frac{\pi a^2}{2}$ (semicircle) — then check any remaining piece by geometry.