Area & Arc Length
in Polar Coordinates

An interactive guide to understanding, deriving, and computing polar area and arc length — with live visualizations.

📐 Conceptual Understanding 🔢 Formula Derivation 🎮 Interactive Visualizations 📝 Worked Examples
1
Polar Coordinates — Quick Review
What is a Polar Coordinate?

Instead of describing a point by its horizontal and vertical distances \((x,y)\), polar coordinates describe it by a distance from the origin \(r\) and an angle from the positive \(x\)-axis \(\theta\).


The conversion between the two systems is:

$$x = r\cos\theta \qquad y = r\sin\theta$$ $$r^2 = x^2 + y^2 \qquad \tan\theta = \frac{y}{x}$$

A polar curve is expressed as \(r = f(\theta)\): the radius changes as the angle sweeps around.

Point:

Try it: Drag the sliders to move the point around the polar plane. Notice how \(r\) sets the distance and \(\theta\) sets the direction.

2
Area in Polar Coordinates
2.1 — Understanding: Why Sectors?

In Cartesian coordinates we slice regions into thin rectangles. In polar coordinates the natural slice is a thin sector (a pie-slice shape), because the coordinates sweep angularly.


A sector with a tiny angular width \(d\theta\) and radius \(r\) has area:

$$dA = \frac{1}{2}r^2\,d\theta$$

This is just the formula for a circle sector \(A=\tfrac{1}{2}r^2\theta\) with \(\theta\to d\theta\). Adding up all infinitesimal sectors from \(\alpha\) to \(\beta\) gives the total area.


The interactive below shows this Riemann-sum approximation. Increase \(n\) to watch the sectors converge to the true area.

Approx Area (sectors)
Exact Area (integral)
Error
2.2 — The Formula
$$\boxed{A = \frac{1}{2}\int_{\alpha}^{\beta} \bigl[r(\theta)\bigr]^2\,d\theta}$$
where \(r(\theta)\) is the polar curve, \(\alpha\) and \(\beta\) are the starting and ending angles of the region being enclosed.
  • For a full curve that closes on itself, integrate over the full period (e.g.\ \(0\) to \(2\pi\) for a cardioid).
  • For a petal of a rose, find the angles where \(r=0\) and integrate between them.
  • For the area between two curves, use \(A=\frac{1}{2}\int_\alpha^\beta(r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta\).
2.3 — Derivation (step by step)
Expand full derivation

Set up a Riemann sum. Divide \([\alpha,\beta]\) into \(n\) equal sub-intervals, each of angular width \(\Delta\theta = \tfrac{\beta-\alpha}{n}\). Choose a sample angle \(\theta_i^*\) in each sub-interval.

Area of one sector. The \(i\)-th sector is nearly a circular sector of radius \(r_i^* = r(\theta_i^*)\) and angle \(\Delta\theta\). A sector of a full circle of radius \(R\) and angle \(\phi\) has area \(\frac{1}{2}R^2\phi\), so:

$$\Delta A_i \approx \frac{1}{2}\bigl[r(\theta_i^*)\bigr]^2\,\Delta\theta$$

Sum all sectors.

$$A \approx \sum_{i=1}^{n} \frac{1}{2}\bigl[r(\theta_i^*)\bigr]^2\,\Delta\theta$$

Take the limit \(n\to\infty\). As \(\Delta\theta\to 0\) the sum becomes a definite integral by the definition of the Riemann integral:

$$A = \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{2}\bigl[r(\theta_i^*)\bigr]^2\Delta\theta = \frac{1}{2}\int_{\alpha}^{\beta}\bigl[r(\theta)\bigr]^2\,d\theta \qquad \blacksquare$$
Why \(\tfrac{1}{2}\)? It comes from the area of a circular sector \(A_\text{sector}=\tfrac{1}{2}r^2\theta\), which itself comes from \(\tfrac{\theta}{2\pi}\cdot\pi r^2\). The factor of \(\tfrac{1}{2}\) is always present and easy to forget!
2.4 — Interactive Area Calculator

Select a curve and drag the angle sliders to define the integration region. The shaded area and its numerical value update in real time.

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Area
3
Arc Length in Polar Coordinates
3.1 — Understanding: Tiny Chord Lengths

Arc length is computed by summing infinitesimal chord lengths along the curve. In polar coordinates the position vector \(\mathbf{r}\) traces a path as \(\theta\) changes. Converting to parametric form:

$$x(\theta) = r(\theta)\cos\theta, \qquad y(\theta) = r(\theta)\sin\theta$$

The arc length formula for parametric curves gives us exactly the polar arc length formula after simplification (see derivation below).

Key idea: The infinitesimal arc element \(ds\) has two contributions — the radial change \(dr\) as the curve moves in/out, and the tangential displacement \(r\,d\theta\) as the angle sweeps. These combine via the Pythagorean theorem: \(ds = \sqrt{dr^2 + r^2\,d\theta^2}\).
3.2 — The Formula
$$\boxed{L = \int_{\alpha}^{\beta}\sqrt{\,r(\theta)^2 + \left(\frac{dr}{d\theta}\right)^{\!2}}\;d\theta}$$
where \(\dfrac{dr}{d\theta} = r'(\theta)\) is the derivative of the radial function.
  • Both \(r^2\) and \((r')^2\) must be non-negative (they always are), so the integrand is always real.
  • For a circle \(r=a\), we get \(r'=0\) and \(L = \int_0^{2\pi}a\,d\theta = 2\pi a\) ✓
  • This integral often lacks a closed form; numerical methods (like Simpson's rule) are frequently needed.
3.3 — Derivation (step by step)
Expand full derivation

Parametric arc length. For any parametric curve \((x(\theta),y(\theta))\), the arc length from \(\alpha\) to \(\beta\) is:

$$L = \int_{\alpha}^{\beta}\sqrt{\left(\frac{dx}{d\theta}\right)^{\!2}+\left(\frac{dy}{d\theta}\right)^{\!2}}\;d\theta$$

Differentiate the conversion formulas. Using \(x=r\cos\theta\) and \(y=r\sin\theta\) with the product rule:

$$\frac{dx}{d\theta} = r'\cos\theta - r\sin\theta, \qquad \frac{dy}{d\theta} = r'\sin\theta + r\cos\theta$$

Square and add.

$$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2$$ $$= (r')^2\cos^2\theta - 2rr'\cos\theta\sin\theta + r^2\sin^2\theta$$ $$\quad + (r')^2\sin^2\theta + 2rr'\sin\theta\cos\theta + r^2\cos^2\theta$$

The cross terms cancel. Group the remaining terms:

$$= (r')^2(\cos^2\theta+\sin^2\theta) + r^2(\sin^2\theta+\cos^2\theta) = (r')^2 + r^2$$

Substitute back.

$$L = \int_{\alpha}^{\beta}\sqrt{r^2+(r')^2}\;d\theta = \int_{\alpha}^{\beta}\sqrt{r(\theta)^2+\left(\frac{dr}{d\theta}\right)^{\!2}}\;d\theta \qquad \blacksquare$$
Geometric intuition: At each angle, the infinitesimal arc element satisfies \(ds^2 = dr^2 + (r\,d\theta)^2\). Dividing by \(d\theta^2\): \(\left(\dfrac{ds}{d\theta}\right)^2 = \left(\dfrac{dr}{d\theta}\right)^2 + r^2\). This is just the Pythagorean theorem on the infinitesimal right triangle with legs \(dr\) (radial) and \(r\,d\theta\) (tangential).
3.4 — Interactive Arc Length Visualizer

The highlighted segment shows the arc between angles \(\alpha\) and \(\beta\). The computed length is shown below.

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Arc Length L
Comparing Area and Arc Length Elements
Area Element \(dA\)
$$dA = \frac{1}{2}r^2\,d\theta$$

Shape: thin circular sector (pie-slice) of radius \(r\) and angle \(d\theta\)

Arc Element \(ds\)
$$ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^{\!2}}\,d\theta$$

Shape: hypotenuse of a right triangle with legs \(r\,d\theta\) and \(dr\)

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Worked Examples
Area Examples
Area inside the cardioid \(r = 1+\cos\theta\)
Setup

The cardioid completes one full revolution for \(\theta\in[0,2\pi]\).

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

Use \(\cos^2\theta = \tfrac{1+\cos 2\theta}{2}\):

$$= \frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}$$
Integrate term by term
$$A = \frac{1}{2}\left[\frac{3}{2}\theta + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi}$$ $$= \frac{1}{2}\cdot\frac{3}{2}\cdot 2\pi = \boxed{\frac{3\pi}{2} \approx 4.712}$$
Area of one petal of \(r = \cos(2\theta)\)
Find petal limits

One petal lies where \(r\ge 0\) in \([-\pi/4,\pi/4]\) (the rightward petal).

$$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4}\cos^2(2\theta)\,d\theta$$
Use half-angle identity
$$\cos^2(2\theta) = \frac{1+\cos 4\theta}{2}$$ $$A = \frac{1}{2}\cdot\frac{1}{2}\int_{-\pi/4}^{\pi/4}(1+\cos 4\theta)\,d\theta$$ $$= \frac{1}{4}\left[\theta + \frac{\sin 4\theta}{4}\right]_{-\pi/4}^{\pi/4} = \frac{1}{4}\cdot\frac{\pi}{2} = \boxed{\frac{\pi}{8} \approx 0.393}$$
Area between two curves
Inside \(r=2\), outside \(r=1+\cos\theta\)

Find intersection: \(2 = 1+\cos\theta \Rightarrow \theta = \pm\pi/2\).

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

Expanding and integrating:

$$= \frac{1}{2}\int_{-\pi/2}^{\pi/2}\!\!\!\left(\frac{5}{2}-2\cos\theta-\frac{\cos 2\theta}{2}\right)d\theta = \boxed{\frac{5\pi}{4}-2 \approx 1.927}$$
Arc Length Examples
Arc length of the circle \(r = 3\)
Setup

\(r = 3\), \(r' = 0\), so:

$$L = \int_0^{2\pi}\sqrt{9+0}\,d\theta = 3\cdot 2\pi = \boxed{6\pi}$$

As expected: circumference of a circle of radius 3.

Arc length of the cardioid \(r = 1+\cos\theta\)
Find \(r'\)
$$r' = -\sin\theta$$
Form the integrand
$$r^2+(r')^2 = (1+\cos\theta)^2+\sin^2\theta$$ $$= 1+2\cos\theta+\cos^2\theta+\sin^2\theta = 2+2\cos\theta$$

Use \(2+2\cos\theta = 4\cos^2(\theta/2)\):

$$\sqrt{r^2+(r')^2} = 2\left|\cos\tfrac{\theta}{2}\right|$$
Integrate
$$L = \int_0^{2\pi}2\left|\cos\tfrac{\theta}{2}\right|\,d\theta = 2\int_0^{2\pi}\cos\tfrac{\theta}{2}\,d\theta = 2\Big[2\sin\tfrac{\theta}{2}\Big]_0^{\pi}$$ $$= 4\cdot 1\cdot 2 = \boxed{8}$$
Arc length of the Archimedean spiral \(r = \theta\) on \([0,2\pi]\)
Setup

\(r=\theta\), \(r'=1\):

$$L = \int_0^{2\pi}\sqrt{\theta^2+1}\,d\theta$$
Evaluate (no closed form in elementary functions)

Use the formula \(\int\sqrt{u^2+1}\,du = \tfrac{1}{2}(u\sqrt{u^2+1}+\sinh^{-1}u)+C\):

$$L = \frac{1}{2}\Big[\theta\sqrt{\theta^2+1}+\ln(\theta+\sqrt{\theta^2+1})\Big]_0^{2\pi}$$ $$\approx \boxed{21.256}$$
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Polar Graph Builder
Build & Compare Polar Curves

Use the clickable keyboard to define curves — no typing needed. Add multiple curves to the same graph, reorder them, then compute the area between any two.

Curves
Active Curve Expression
r = _

Click a curve above to edit it. Build its expression with the keyboard below.

Expression Keyboardθ is the angle variable
Quick Presets
Area Between Two Curves
Outer curve (larger r)
Inner curve (smaller r)
Area between curves — select two curves —