In short

When a curve is given parametrically as x = x(t), y = y(t), the enclosed area is \int y\,dx = \int y(t)\,x'(t)\,dt. When a curve is given in polar form as r = f(\theta), the area it sweeps out is \frac{1}{2}\int r^2\,d\theta. Both formulas come from the same strip-slicing idea you already know — adapted to the new coordinate system.

A point moves along a curve. At time t, its position is x(t) = a\cos t, y(t) = b\sin t. As t runs from 0 to 2\pi, the point traces an ellipse — but you never write down a single equation y = f(x). The curve is described parametrically: each coordinate is a separate function of the parameter t.

How do you find the area enclosed by this ellipse?

You cannot just integrate y\,dx the usual way, because y is not written as a function of x. But you can express everything in terms of t. Since dx = x'(t)\,dt = -a\sin t\,dt, the integral becomes

A = \int y\,dx = \int_0^{2\pi} (b\sin t)(-a\sin t)\,dt = -ab\int_0^{2\pi} \sin^2 t\,dt

Using \sin^2 t = (1 - \cos 2t)/2:

A = -ab \cdot \frac{1}{2}\int_0^{2\pi} (1 - \cos 2t)\,dt = -ab \cdot \frac{1}{2} \cdot 2\pi = -\pi ab

The negative sign appears because as t goes from 0 to 2\pi, the curve is traversed counterclockwise while x first decreases and then increases. Taking the absolute value:

A = \pi ab

This is the area of an ellipse with semi-axes a and b. When a = b = r, you get \pi r^2 — the circle area formula, derived from parametric integration.

Area using parametric curves

Area enclosed by a parametric curve

If a closed curve is given by x = x(t), y = y(t) for t \in [\alpha, \beta], and the curve is traversed once, the enclosed area is

A = \left|\int_\alpha^\beta y(t)\,x'(t)\,dt\right| = \left|\int_\alpha^\beta x(t)\,y'(t)\,dt\right|

Equivalently, using the symmetric form:

A = \frac{1}{2}\left|\int_\alpha^\beta \bigl[x(t)\,y'(t) - y(t)\,x'(t)\bigr]\,dt\right|

Reading the definition. The first form, \int y\,dx, is the area computed by vertical strips, rewritten in the parameter t using dx = x'(t)\,dt. The second form, \int x\,dy, uses horizontal strips. The symmetric form averages both and is sometimes more convenient.

The absolute value handles the sign that depends on the direction of traversal: counterclockwise gives one sign, clockwise the opposite.

When the curve is not closed. If the curve runs from point A to point B (not a closed loop), \int_\alpha^\beta y(t)\,x'(t)\,dt gives the signed area between the curve, the x-axis, and the vertical lines through A and B. The interpretation is the same as for \int_a^b f(x)\,dx — just expressed parametrically.

Polar coordinates and area

Some curves are naturally described in polar coordinates. A point at distance r from the origin, making angle \theta with the positive x-axis, has polar coordinates (r, \theta). A curve given as r = f(\theta) sweeps out area as \theta varies.

The key insight: in polar coordinates, the natural "strip" is not a rectangle — it is a thin sector (a pizza-slice shape). A sector with radius r and angle d\theta has area \frac{1}{2}r^2\,d\theta (this follows from the formula for the area of a circular sector: \frac{1}{2}r^2\theta).

Area in polar coordinates

The area enclosed by the polar curve r = f(\theta) from \theta = \alpha to \theta = \beta is

A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta = \frac{1}{2}\int_\alpha^\beta [f(\theta)]^2\,d\theta

Reading the definition. As \theta sweeps from \alpha to \beta, the radius r = f(\theta) traces out a region. At each angle, the thin sector has area \frac{1}{2}r^2\,d\theta. Integrating adds up all these thin sectors.

A quick sanity check. For a circle of radius R, the polar equation is r = R (constant). The full area is

A = \frac{1}{2}\int_0^{2\pi} R^2\,d\theta = \frac{1}{2}R^2 \cdot 2\pi = \pi R^2

Correct — as it must be.

Area of a cardioid

The cardioid r = 1 + \cos\theta is a heart-shaped curve that appears frequently in JEE problems. It traces one full loop as \theta runs from 0 to 2\pi.

A = \frac{1}{2}\int_0^{2\pi} (1 + \cos\theta)^2\,d\theta

Expand: (1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta.

A = \frac{1}{2}\int_0^{2\pi} \left(1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2}\right)\,d\theta = \frac{1}{2}\int_0^{2\pi} \left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)\,d\theta

Over a full period, \int_0^{2\pi}\cos\theta\,d\theta = 0 and \int_0^{2\pi}\cos 2\theta\,d\theta = 0. So:

A = \frac{1}{2} \cdot \frac{3}{2} \cdot 2\pi = \frac{3\pi}{2}

The cardioid encloses an area of 3\pi/2. Compared to the circle r = 2 (which has area 4\pi), the cardioid fills about 37.5\% of the circle that just contains it — the indentation on one side removes a substantial chunk.

The cardioid $r = 1 + \cos\theta$, shown in Cartesian form. It has a cusp at the origin and extends to $x = 2$ on the right. The enclosed area is $3\pi/2$.

Variable area problems

A different type of problem: the area itself is not a fixed number, but depends on a parameter.

Setup. Consider the family of regions bounded by y = x^2 and y = a for different values of a > 0. The horizontal line y = a cuts the parabola at x = \pm\sqrt{a}, creating a region that grows as a increases. Express the enclosed area as a function of a.

The parabola y = x^2 is below the line y = a for -\sqrt{a} < x < \sqrt{a}. The area between them is

A(a) = \int_{-\sqrt{a}}^{\sqrt{a}} (a - x^2)\,dx

By symmetry, this is twice the integral from 0 to \sqrt{a}:

A(a) = 2\int_0^{\sqrt{a}} (a - x^2)\,dx = 2\left[ax - \frac{x^3}{3}\right]_0^{\sqrt{a}} = 2\left(a\sqrt{a} - \frac{a\sqrt{a}}{3}\right) = 2 \cdot \frac{2a\sqrt{a}}{3} = \frac{4}{3}a^{3/2}

The area grows as a^{3/2} — faster than linearly, because both the width and the height of the region increase with a.

Rate of change of area. Since A(a) = \frac{4}{3}a^{3/2}, the rate at which the area changes with respect to a is

\frac{dA}{da} = \frac{4}{3} \cdot \frac{3}{2}a^{1/2} = 2\sqrt{a}

At a = 4, the area is growing at a rate of 2\sqrt{4} = 4 square units per unit increase in a. Notice that 2\sqrt{a} is exactly the width of the region at height a — the area increases by (width \times height-increase), which makes geometric sense.

The parabola $y = x^2$ cut by horizontal lines $y = 2$ and $y = 4$. The enclosed region grows both wider and taller as $a$ increases. At $a = 2$, the area is $\frac{4}{3}(2)^{3/2} = \frac{8\sqrt{2}}{3} \approx 3.77$. At $a = 4$, the area is $\frac{4}{3}(4)^{3/2} = \frac{32}{3} \approx 10.67$.

Two complete worked examples

Example 1: Area enclosed by one arch of a cycloid

A cycloid is the curve traced by a point on the rim of a rolling wheel. If the wheel has radius r, the parametric equations are x = r(t - \sin t), y = r(1 - \cos t). One arch corresponds to t \in [0, 2\pi]. Find the area under one arch.

Step 1. Write the area integral in parametric form.

A = \int y\,dx = \int_0^{2\pi} r(1 - \cos t) \cdot r(1 - \cos t)\,dt = r^2\int_0^{2\pi}(1 - \cos t)^2\,dt

Why: here dx = x'(t)\,dt = r(1 - \cos t)\,dt. The integrand becomes y(t) \cdot x'(t) = r(1-\cos t) \cdot r(1-\cos t) = r^2(1-\cos t)^2.

Step 2. Expand (1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t.

A = r^2\int_0^{2\pi}\left(1 - 2\cos t + \frac{1 + \cos 2t}{2}\right)dt = r^2\int_0^{2\pi}\left(\frac{3}{2} - 2\cos t + \frac{\cos 2t}{2}\right)dt

Why: use the identity \cos^2 t = (1 + \cos 2t)/2 to eliminate the squared trig term.

Step 3. Integrate term by term over [0, 2\pi].

\int_0^{2\pi}\frac{3}{2}\,dt = 3\pi, \quad \int_0^{2\pi}2\cos t\,dt = 0, \quad \int_0^{2\pi}\frac{\cos 2t}{2}\,dt = 0

Why: cosine integrates to zero over a full period. Only the constant term survives.

Step 4. Combine.

A = r^2 \cdot 3\pi = 3\pi r^2

Why: the area under one arch of a cycloid is exactly three times the area of the generating circle. This is a beautiful result — the shape traced by a rolling circle captures exactly three circles' worth of area.

Result: The area under one arch of a cycloid is 3\pi r^2.

One arch of the cycloid with $r = 1$. The curve starts and ends on the $x$-axis and reaches a maximum height of $2r = 2$ at the midpoint. The area under this arch is $3\pi \approx 9.42$, exactly three times the area of the generating circle ($\pi r^2 = \pi$).

The result 3\pi r^2 is one of the most surprising in classical geometry. A circle of radius r has area \pi r^2. Rolling it along a flat surface traces an arch whose area is exactly three times that — no more, no less. The factor of 3 is not approximate; it is exact.

Example 2: Area enclosed by one petal of the rose curve $r = \cos 2\theta$

The polar curve r = \cos 2\theta is a four-petalled rose. Find the area of one petal.

Step 1. Determine the range of \theta for one petal. A petal exists where r \geq 0. Since \cos 2\theta \geq 0 when -\pi/4 \leq \theta \leq \pi/4, the right-pointing petal corresponds to \theta \in [-\pi/4, \pi/4].

Why: when \cos 2\theta < 0, the radius is negative, which in polar coordinates means the point is plotted in the opposite direction — that gives a different petal. Each petal spans a \pi/2 range of \theta.

Step 2. Set up the polar area integral.

A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} (\cos 2\theta)^2\,d\theta

Why: the polar area formula is \frac{1}{2}\int r^2\,d\theta, and here r = \cos 2\theta.

Step 3. Simplify using \cos^2 2\theta = \frac{1 + \cos 4\theta}{2}.

A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1 + \cos 4\theta}{2}\,d\theta = \frac{1}{4}\int_{-\pi/4}^{\pi/4} (1 + \cos 4\theta)\,d\theta

Why: the double-angle identity converts the squared cosine into a form you can integrate directly.

Step 4. Evaluate.

A = \frac{1}{4}\left[\theta + \frac{\sin 4\theta}{4}\right]_{-\pi/4}^{\pi/4} = \frac{1}{4}\left[\left(\frac{\pi}{4} + \frac{\sin\pi}{4}\right) - \left(-\frac{\pi}{4} + \frac{\sin(-\pi)}{4}\right)\right]
= \frac{1}{4}\left[\frac{\pi}{4} + 0 + \frac{\pi}{4} - 0\right] = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}

Result: The area of one petal is \dfrac{\pi}{8}.

The four-petalled rose $r = \cos 2\theta$ (shown in Cartesian form as $(x^2+y^2)^2 = x^2 - y^2$, which only shows two petals since this implicit form captures $r^2 = \cos 2\theta$). Each petal has area $\pi/8$, so the total area of all four petals is $\pi/2$ — half the area of the unit circle.

The total area of all four petals is 4 \times \pi/8 = \pi/2. The unit circle (r = 1) has area \pi, so the rose fills exactly half the circle. Each petal is a thin, elegant shape — \pi/8 \approx 0.393 square units, barely more than a third of a square unit.

Common confusions

Going deeper

If you came here to learn parametric and polar area formulas, you have them. The rest of this section looks at why the formulas work and how they connect to deeper ideas.

Why the polar area formula has r^2

In Cartesian coordinates, a thin strip is a rectangle: width dx, height y, area y\,dx. In polar coordinates, a thin sector is a triangle-like wedge: two sides of length r, included angle d\theta. The area of a triangle with two sides r and included angle d\theta is \frac{1}{2}r \cdot r \cdot \sin(d\theta) \approx \frac{1}{2}r^2\,d\theta (since \sin(d\theta) \approx d\theta for small angles). That is where the \frac{1}{2}r^2 comes from.

Green's theorem connection

The parametric area formula A = \frac{1}{2}\oint(x\,dy - y\,dx) is a special case of Green's theorem from multivariable calculus. Green's theorem says that a line integral around a closed curve equals a double integral over the enclosed region. When you choose the right vector field, the double integral simplifies to just the area. This is the same mathematics, viewed from a higher altitude.

Area between two polar curves

If you have two polar curves r_1 = f(\theta) and r_2 = g(\theta) with f(\theta) \geq g(\theta), the area of the region between them from \theta = \alpha to \theta = \beta is

A = \frac{1}{2}\int_\alpha^\beta \bigl[f(\theta)^2 - g(\theta)^2\bigr]\,d\theta

This is the polar analogue of "top minus bottom" in Cartesian coordinates: each thin sector's area is the outer sector minus the inner sector.

Where this leads next