Area and arc length in polar coordinates · Calculus II: Integrals & Series

Arc length in polar form is a direct application of §6.1's parametric formula — nothing new to derive. Area is genuinely different: polar regions are naturally swept out by rotating a ray, not by stacking vertical strips, so §5.0's "top minus bottom" picture doesn't apply at all. This closing lesson of the module builds area from scratch, the polar way.

Area: summing thin pie slices

A vertical strip approximates area with a rectangle. In polar coordinates, the natural small piece is a thin circular sector — a pie slice spanning angle d\theta at radius r=f(\theta). A full circle sector of angle \theta and radius r has area \frac12r^2\theta (this is where the area of a full circle, A=\pi r^2 from §0.1, comes from: \theta=2\pi gives \frac12r^2(2\pi)=\pi r^2). For a sector of infinitesimal angle d\theta:

$dA=\frac12\big[f(\theta)\big]^2d\theta$

Summing sectors from \theta=\alpha to \theta=\beta:

$A=\frac12\int_\alpha^\beta\big[f(\theta)\big]^2d\theta$

The exponent tells you this is fundamentally a "disk method" idea (§5.1) rather than a "strip" idea — area grows with r^2, the same way a circle's area does, because each pie slice's area itself already involves r^2 before any integration happens.

Area between two polar curves, outer r=f(\theta) and inner r=g(\theta) (both measured from the same origin, over the same angle range) works exactly like §5.1's washers — outer squared minus inner squared, not the squared difference:

$A=\frac12\int_\alpha^\beta\Big(\big[f(\theta)\big]^2-\big[g(\theta)\big]^2\Big)d\theta$

Finding the bounds is often the hard part. Because polar representations aren't unique (§6.2), the angle range that traces out "the region" isn't always obvious — sketch the curve, or solve f(\theta)=g(\theta) for intersections, before setting up the integral.

Arc length: the parametric formula, specialized

A polar curve is parametric with x=f(\theta)\cos\theta, y=f(\theta)\sin\theta (§6.2). Plugging straight into §6.1's arc-length formula and simplifying with the product rule and the Pythagorean identity:

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

$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=\big[f'(\theta)\big]^2+\big[f(\theta)\big]^2$

— every cross term cancels (expand it out; the \sin\theta\cos\theta pieces are opposite in sign between the two squared expressions), leaving:

$L=\int_\alpha^\beta\sqrt{\big[f(\theta)\big]^2+\big[f'(\theta)\big]^2}\,d\theta$

For the special case r=a (a plain circle), f'(\theta)=0, and this correctly reduces to L=\int_0^{2\pi}a\,d\theta=2\pi a — ordinary circumference, confirming the formula against a case you can check by eye.

Doing it in Python

Area enclosed by the cardioid r=1+\cos\theta, swept over a full 2\pi:

import sympy as sp

theta = sp.Symbol('theta')
r = 1 + sp.cos(theta)

A = sp.Rational(1, 2) * sp.integrate(r**2, (theta, 0, 2 * sp.pi))
print(f"area enclosed by the cardioid = {A}")

Area of a single petal of the rose r=\cos(3\theta) — the petal near \theta=0 spans exactly \theta\in\left[-\frac\pi6,\frac\pi6\right], the range where \cos(3\theta)\ge0:

import sympy as sp

theta = sp.Symbol('theta')
r = sp.cos(3 * theta)

petal_area = sp.Rational(1, 2) * sp.integrate(r**2, (theta, -sp.pi/6, sp.pi/6))
print(f"area of one petal = {petal_area}")
print(f"area of all 3 petals = {3 * petal_area}")

Arc length of the full cardioid, using the half-angle identity to resolve the square root — the same technique the cycloid needed in §6.1:

import sympy as sp

theta = sp.Symbol('theta')
r = 1 + sp.cos(theta)
rp = sp.diff(r, theta)

integrand = sp.simplify(sp.sqrt(r**2 + rp**2))
print(f"integrand simplifies to: {integrand}")
# 2 + 2cos(theta) = 4cos^2(theta/2), so sqrt(...) = 2|cos(theta/2)|
# split at theta=pi where cos(theta/2) changes sign
L = (sp.integrate(2 * sp.cos(theta / 2), (theta, 0, sp.pi))
     + sp.integrate(-2 * sp.cos(theta / 2), (theta, sp.pi, 2 * sp.pi)))
print(f"arc length of the full cardioid = {L}")

Worked example

Find the area enclosed by one petal of the rose r=\cos(3\theta).

First, find where the petal begins and ends — where r=0:

$\cos(3\theta)=0\ \Longrightarrow\ 3\theta=\pm\frac\pi2\ \Longrightarrow\ \theta=\pm\frac\pi6$

Between \theta=-\frac\pi6 and \theta=\frac\pi6, \cos(3\theta)\ge0 — one full petal, traced out and back to the origin.

$A=\frac12\int_{-\pi/6}^{\pi/6}\cos^2(3\theta)\,d\theta$

Half-angle identity (§4.7): \cos^2(3\theta)=\frac{1+\cos(6\theta)}2.

$A=\frac14\int_{-\pi/6}^{\pi/6}\big(1+\cos6\theta\big)d\theta=\frac14\left[\theta+\frac{\sin6\theta}6\right]_{-\pi/6}^{\pi/6}$

At \theta=\pm\frac\pi6: \sin(6\cdot\pm\frac\pi6)=\sin(\pm\pi)=0, so the sine terms vanish entirely.

$A=\frac14\left[\frac\pi6-\left(-\frac\pi6\right)\right]=\frac14\cdot\frac\pi3=\boxed{\frac\pi{12}}$

Sanity check. One petal fits inside a circle of radius 1 (since |\cos(3\theta)|\le1 always), swept over an angular range of \frac\pi3 out of the full 2\pi — a full unit disk has area \pi, and a petal covering roughly \frac1{6} of that angular range but pinched to a point at both ends should be noticeably less than \frac\pi6\approx0.524. \frac\pi{12}\approx0.262 fits. ✓

Your turn

1. Find the area enclosed by the circle r=2\sin\theta (swept over \theta\in[0,\pi] — check that this range alone already covers the whole circle by testing the endpoint radii).

2. Find the arc length of the spiral r=\theta for \theta\in[0,\pi].

3. True or false: sweeping \theta from 0 to 2\pi always traces the full region enclosed by a polar curve r=f(\theta) exactly once.

Solutions

1. At \theta=0: r=0. At \theta=\frac\pi2: r=2 (the far side of the circle). At \theta=\pi: r=0 again — the curve returns to the origin, confirming [0,\pi] already closes the loop (sweeping to 2\pi would simply retrace it, since \sin\theta<0 for \theta\in(\pi,2\pi) and a negative r there reflects back onto the same circle).

$A=\frac12\int_0^\pi(2\sin\theta)^2d\theta=2\int_0^\pi\sin^2\theta\,d\theta=2\int_0^\pi\frac{1-\cos2\theta}2d\theta=\left[\theta-\frac{\sin2\theta}2\right]_0^\pi=\boxed\pi$

Matches the ordinary area of a circle of radius 1 (this curve has diameter 2, per §6.2's r=a\sin\theta identification): \pi(1)^2=\pi. ✓

2. f(\theta)=\theta, f'(\theta)=1:

$L=\int_0^\pi\sqrt{\theta^2+1}\,d\theta$

This is the same integral form as §5.3's \int\sqrt{1+x^2}dx (a trig substitution, §4.8), with the standard antiderivative \frac\theta2\sqrt{\theta^2+1}+\frac12\ln\left(\theta+\sqrt{\theta^2+1}\right):

$L=\left[\frac\theta2\sqrt{\theta^2+1}+\frac12\ln\!\left(\theta+\sqrt{\theta^2+1}\right)\right]_0^\pi=\boxed{\frac\pi2\sqrt{\pi^2+1}+\frac12\ln\!\left(\pi+\sqrt{\pi^2+1}\right)}$

3. False. Some curves need less than a full 2\pi to close (like problem 1's circle, done by \theta=\pi, or the rose petals, each done in a range of \frac\pi3); others, like a rose with an even number of petals or a curve involving \cos(n\theta) with certain n, may need \theta to run all the way to 4\pi before every petal has been traced — and some curves retrace themselves entirely within a smaller range, making a naive [0,2\pi] integral double-count area. Always check where r actually returns to its starting behavior before fixing the bounds.

Check yourself in code

Compute the area enclosed by the cardioid r=1+\cos\theta and the area of one petal of the rose r=\cos(3\theta).

Print exactly this:

cardioid area = 3*pi/2
one rose petal area = pi/12

import sympy as sp

theta = sp.Symbol('theta')

r1 = 1 + sp.cos(theta)
A1 = sp.Rational(1, 2) * sp.integrate(r1**2, (theta, 0, 2 * sp.pi))
print("cardioid area = ...")

r2 = sp.cos(3 * theta)
A2 = sp.Rational(1, 2) * sp.integrate(r2**2, (theta, -sp.pi/6, sp.pi/6))
print("one rose petal area = ...")

import sympy as sp

theta = sp.Symbol('theta')

r1 = 1 + sp.cos(theta)
A1 = sp.Rational(1, 2) * sp.integrate(r1**2, (theta, 0, 2 * sp.pi))
print(f"cardioid area = {A1}")

r2 = sp.cos(3 * theta)
A2 = sp.Rational(1, 2) * sp.integrate(r2**2, (theta, -sp.pi/6, sp.pi/6))
print(f"one rose petal area = {A2}")

Polar area sums thin pie slices, A=\frac12\int r^2\,d\theta — a disk-method idea rather than a strip-method one, since a sector's area already carries r^2 before any integration — while polar arc length is nothing but §6.1's parametric formula with the cross terms cancelling to leave L=\int\sqrt{r^2+(dr/d\theta)^2}\,d\theta. The recurring difficulty in this lesson isn't the calculus, which is routine, but the bookkeeping: because polar coordinates aren't unique, finding the correct angle range to sweep takes real care, usually a sketch or a check of where r returns to zero.

That's this module's full toolkit for curves beyond y=f(x) — parametrized by a free parameter or by angle and radius, each with its own derivative, arc length, and area formulas, all traceable back to the same Riemann-sum argument this course started with in Module 4. Next: sequences and series — where the subject turns from continuous curves to infinite sums, starting the second half of the course.

25. Area and arc length in polar coordinates