Finding the Area of a Polar Region or the Area Bounded by a Single Polar Curve

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Finding the Area of a Polar Region

Area is swept, not stacked: thin sectors of radius r and angle dθ.

key idea$A=\frac{1}{2}\int_{\alpha}^{\beta} r^{2}\,d\theta$

Picture the region

Polar area is swept, not stacked. The element is a thin circular sector of radius $r$ and angle $d\theta$, whose area is $\tfrac12r^{2}d\theta$ — the sector formula from geometry. Adding those wedges as $\theta$ turns is the integral.

a wedge of the cardioid $r=2+2\cos\theta$

one petal of the rose $r=3\cos2\theta$

Left: the wedge swept from $\theta=0$ to $\theta=\dfrac{\pi}{3}$ on a cardioid, bounded by two amber rays. Right: one petal of a four-petal rose, swept between the two angles where $r=0$. In both cases the limits are angles, and finding them is most of the work.

Setting up the integral

Sweeping out a polar area

Sketch the curve well enough to see which angles bound the region you want.
Find the limits: for a full closed curve, one complete revolution; for a petal or loop, solve $r=0$ and take two consecutive solutions.
Write $\dfrac12\displaystyle\int_{\alpha}^{\beta}\left[f(\theta)\right]^{2}d\theta$. The $\tfrac12$ is part of the formula, not an afterthought.
Expand the square before integrating. Squaring a sum such as $2+2\cos\theta$ produces a $\cos^{2}$ term that needs the power-reduction identity.
Watch for double counting: sweeping past $2\pi$, or over an interval where the curve retraces, adds the same area twice.

Region	Limits	Why
one petal of $r=a\cos(n\theta)$, $n$ odd	consecutive zeros of $r$	the petal starts and ends at the pole
one petal of $r=a\cos(n\theta)$, $n$ even	consecutive zeros of $r$	same rule — but there are $2n$ petals
a full cardioid	$0$ to $2\pi$	one revolution traces it exactly once
the inner loop of a limaçon	the two angles where $r=0$	the loop is traced between them
a circle $r=a\cos\theta$	$-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$	a full revolution would trace it twice

Row 5 is the classic double count. $r=a\cos\theta$ completes its circle in a half-turn, so integrating over $[0,2\pi]$ returns exactly twice the correct area.

The elementa sector, area $\tfrac12r^{2}d\theta$

The limitsangles, usually from $r=0$

Before integratingexpand and power-reduce

Worked examples

EX 1

One petal of a rose

limits from $r=0$

Find the area of one petal of $r=3\cos2\theta$.

Find where the petal begins and ends

$3\cos2\theta=0$ gives $2\theta=\pm\dfrac{\pi}{2}$, so $\theta=\pm\dfrac{\pi}{4}$.

Set up

$A=\dfrac12\displaystyle\int_{-\pi/4}^{\pi/4}9\cos^{2}2\theta\,d\theta$.

Power-reduce

$\cos^{2}2\theta=\dfrac{1+\cos4\theta}{2}$, so the integrand is $\dfrac92\left(1+\cos4\theta\right)$ after the outer $\tfrac12$ gives $\dfrac94\left(1+\cos4\theta\right)$.

Integrate

$\dfrac94\left[\theta+\dfrac{\sin4\theta}{4}\right]_{-\pi/4}^{\pi/4}=\dfrac94\left[\dfrac{\pi}{2}+0\right]$.

Evaluate

$A=\dfrac{9\pi}{8}$.

EX 2

A full cardioid

one revolution

Find the area enclosed by $r=2+2\cos\theta$.

Choose the limits

The cardioid closes after one full turn, so $\theta$ runs from $0$ to $2\pi$.

Square the radius

$\left(2+2\cos\theta\right)^{2}=4+8\cos\theta+4\cos^{2}\theta$.

Power-reduce the last term

$4\cos^{2}\theta=2+2\cos2\theta$, so the bracket becomes $6+8\cos\theta+2\cos2\theta$.

Integrate over a full turn

$\displaystyle\int_{0}^{2\pi}\cos\theta\,d\theta=0$ and $\displaystyle\int_{0}^{2\pi}\cos2\theta\,d\theta=0$, leaving $6(2\pi)=12\pi$.

Apply the one-half

$A=\dfrac12(12\pi)=6\pi$.

EX 3

Avoiding a double count

the circle $r=4\cos\theta$

Find the area enclosed by $r=4\cos\theta$.

Identify the curve

Converting gives $x^{2}+y^{2}=4x$, a circle of radius $2$ centred at $(2,0)$, so the answer should be $4\pi$.

Find the correct limits

$r=0$ at $\theta=\pm\dfrac{\pi}{2}$, and the circle is complete over that half-turn.

Set up

$A=\dfrac12\displaystyle\int_{-\pi/2}^{\pi/2}16\cos^{2}\theta\,d\theta=4\int_{-\pi/2}^{\pi/2}\left(1+\cos2\theta\right)d\theta$.

Evaluate

$4\left[\theta+\dfrac{\sin2\theta}{2}\right]_{-\pi/2}^{\pi/2}=4\pi$. ✓

See the trap

Integrating over $[0,2\pi]$ would give $8\pi$ — the circle traced twice, once with $r\gt 0$ and once with $r\lt 0$.

EX 4

The inner loop

a limaçon

Set up the area of the inner loop of $r=1+2\cos\theta$.

Find the pole crossings

$1+2\cos\theta=0$ gives $\cos\theta=-\dfrac12$, so $\theta=\dfrac{2\pi}{3}$ and $\theta=\dfrac{4\pi}{3}$.

Identify which arc is the inner loop

Between those angles $r$ is negative, and negative radii plot back through the pole — that is the inner loop.

Set up the integral

$A=\dfrac12\displaystyle\int_{2\pi/3}^{4\pi/3}\left(1+2\cos\theta\right)^{2}d\theta$.

Note that squaring handles the sign

$r^{2}$ is positive even where $r$ is not, so no absolute value is needed.

Evaluate

Expanding and power-reducing gives $A=\pi-\dfrac{3\sqrt3}{2}\approx0.544$.

Watch out

The $\tfrac12$ belongs to the formula

Area is $\dfrac12\int r^{2}d\theta$, from the sector formula. Omitting it doubles every answer.

Square the radius, do not integrate $r$

$\int r\,d\theta$ is not an area and has the wrong units. The element is a sector, not a rectangle.

Find the limits from $r=0$

Petals and loops begin and end at the pole. Guessing the limits from the picture is where most of the error in this topic lives.

A full revolution can double count

Curves like $r=a\cos\theta$ close in half a turn. Check how much of a revolution traces the curve once before setting the limits.

Course map. Topic 9.8 · LO CHA-5.D — calculate areas of regions defined by polar curves using definite integrals. BC-only.