How Calculus Finds the Area of an Ellipse-Shaped Garden

School formulas give the areas of rectangles, triangles and circles, but not of a region bounded by a parabola or a wavy curve. A definite integral adds up thin strips under any curve, turning such areas into a calculation.

This part covers setting up area integrals, areas bounded by lines, circles, parabolas and ellipses, and regions below the axis or split into pieces.

**The area between a curve $y=f(x)\ge 0$, the $x$-axis and the lines $x=a$ and $x=b$ is ${\int }_a^{b}ydx$, and the area between $x=g(y)\ge 0$, the $y$-axis and the lines $y=c$ and $y=d$ is ${\int }_c^{d}xdy$.

Vertical strips.** A thin strip of width $dx$ and height $y$ has area $ydx$; adding all the strips from $a$ to $b$ gives

$$A={\int }_a^{b}ydx$$

Horizontal strips. For a region beside the $y$-axis, strips of height $dy$ and length $x$ give $A={\int }_c^{d}xdy$.

Worked example 1. The area under $y=x^2$ from $x=1$ to $x=3$:

$$A={\int }_1^3{x}^{2}dx={\left[\frac{x^3}{3}\right]}_1^3=9-\frac{1}{3}=\frac{26}{3}\text{ sq units}$$

Worked example 2. The area between $x=y^2$, the $y$-axis and the lines $y=0$ and $y=2$:

$$A={\int }_0^2{y}^{2}dy={\left[\frac{y^3}{3}\right]}_0^2=\frac{8}{3}\text{ sq units}$$

An everyday example. Estimating the area of a curved paddy field by pacing out many narrow parallel strips is exactly how the integral adds up thin rectangles.

The substance. Sketch the region first — the sketch decides whether vertical or horizontal strips give the simpler integral.

**Use symmetry to find the area in one quadrant or half, integrate the curve's equation solved for $y$, and multiply back; this gives $\pi a^2$ for the circle $x^2+y^2=a^2$ and $\pi ab$ for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Worked example 1 — the circle $x^2+y^2=a^2$.** In the first quadrant, $y=\sqrt{a^2-x^2}$:

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

**Worked example 2 — the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$.** Here $y=\frac{3}{4}\sqrt{16-x^2}$:

$$A=4\cdot \frac{3}{4}{\int }_0^4\sqrt{16-x^2}dx=3\cdot \frac{16}{2}\cdot \frac{\pi }{2}=12\pi \text{ sq units}$$

This matches $\pi ab=\pi (4)(3)$.

Worked example 3 — a parabola and a line. The region enclosed by $y^2=4x$ and $x=4$ has upper boundary $y=2\sqrt{x}$:

$$A=2{\int }_0^{4}2\sqrt{x}dx=4{\left[\frac{2}{3}{x}^{3/2}\right]}_0^4=\frac{64}{3}\text{ sq units}$$

An everyday example. **An elliptical flower bed $8$ m long and $6$ m wide** has area $\pi \times 4\times 3\approx 37.7$ m${}^2$ — the ellipse formula at work.

The substance. Symmetry is a shortcut, not an assumption — check that the region really is symmetric before multiplying.

**Where a curve lies below the $x$-axis its integral is negative, so its absolute value gives the area; when a region crosses the axis or changes boundary, split it at those points and add the separate areas.

Below the axis.** If $f(x)\le 0$ on $[a,b]$, then $A=\left|{\int }_a^{b}f(x)dx\right|$.

Worked example 1 — crossing the axis. The area between $y=\sin x$ and the $x$-axis from $0$ to $2\pi$:

$${\int }_0^{\pi }\sin xdx=2{\int }_{\pi }^{2\pi }\sin xdx=-2A=2+|-2|=4\text{ sq units}$$

Integrating straight from $0$ to $2\pi$ would wrongly give $0$.

Worked example 2 — splitting at a meeting point. The region bounded by $y=x$, $y=2-x$ and the $x$-axis; the lines meet at $(1,1)$:

$$A={\int }_0^{1}xdx+{\int }_1^2(2-x)dx=\frac{1}{2}+\frac{1}{2}=1\text{ sq unit}$$

This agrees with a triangle of base $2$ and height $1$.

An everyday example. A shop with a ₹2 lakh profit one month and a ₹2 lakh loss the next shows zero net change, yet ₹4 lakh actually changed hands — just as signed integrals can cancel while the true area adds up.

The substance. Area is always positive — a signed integral measures area only after the negative parts are made positive.