Finding the Area of the Region Bounded by Two Polar Curves

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

Area Bounded by Two Polar Curves

Outer sector minus inner sector — with the intersection angles found before anything else.

key idea$A=\frac{1}{2}\int_{\alpha}^{\beta}\Big[r_{\text{out}}^{2}-r_{\text{in}}^{2}\Big]d\theta$

Picture the region

Two polar curves bound a region the way two Cartesian curves do — by subtraction. Each sector runs from the inner curve out to the outer one, so the element is a difference of two sector areas. Squaring happens first, and the subtraction second.

The cardioid $r=2+2\cos\theta$ against the circle $r=3$, with the wedge between them shaded from $\theta=-\dfrac{\pi}{3}$ to $\dfrac{\pi}{3}$. Over that range the cardioid is outside the circle; past those angles they swap, which is exactly why the intersection angles have to be found first.

Setting up the integral

Two curves, one wedge at a time

Find the intersection angles by setting $r_1(\theta)=r_2(\theta)$ and solving. These are almost always the limits.
Check for intersections at the pole separately. Two curves can both reach the origin without ever sharing a value of $\theta$, so this crossing does not appear in the algebra.
Test one angle inside each range to see which curve is farther out. Substitute; do not rely on the sketch.
Write $\dfrac12\displaystyle\int\left[r_{\text{out}}^{2}-r_{\text{in}}^{2}\right]d\theta$ — square each radius, then subtract.
If the roles swap partway, split the integral at that angle and add the pieces, exactly as with Cartesian regions.

Expression	Correct?	Why
$\dfrac12\displaystyle\int\left(r_{\text{out}}^{2}-r_{\text{in}}^{2}\right)d\theta$	yes	difference of two sectors
$\dfrac12\displaystyle\int\left(r_{\text{out}}-r_{\text{in}}\right)^{2}d\theta$	no	squares a gap that is not a radius
$\dfrac12\displaystyle\int r_{\text{out}}^{2}-\dfrac12\int r_{\text{in}}^{2}$	yes	the same thing, split into two integrals
$\displaystyle\int\left(r_{\text{out}}-r_{\text{in}}\right)d\theta$	no	no $r^{2}$, no $\tfrac12$ — not an area at all

Row 2 is the same error that haunts the washer method, in polar clothing: the area between two sectors is not the area of a sector whose radius is the gap.

four decisions

Set $r_1=r_2$ for the limits — and check the pole separately, because it hides from that equation.
Test an interior angle to decide which curve is outer. The answer can change from one range to the next.
Square each radius before subtracting.
A region that is inside one curve and outside another is a subtraction; a region formed by different curves in different ranges is a sum of pieces.

Worked examples

EX 1

Inside one, outside the other

the standard set-up

Find the area inside $r=3$ and outside $r=2+2\cos\theta$.

Find the intersections

$3=2+2\cos\theta$ gives $\cos\theta=\dfrac12$, so $\theta=\pm\dfrac{\pi}{3}$.

Decide which is outer

At $\theta=\pi$: the circle gives $3$, the cardioid gives $0$. So on $\left[\dfrac{\pi}{3},\dfrac{5\pi}{3}\right]$ the circle is outside.

Set up

$A=\dfrac12\displaystyle\int_{\pi/3}^{5\pi/3}\left[9-\left(2+2\cos\theta\right)^{2}\right]d\theta$.

Expand and power-reduce

$\left(2+2\cos\theta\right)^{2}=6+8\cos\theta+2\cos2\theta$ after reducing, so the bracket is $3-8\cos\theta-2\cos2\theta$.

Evaluate

$\dfrac12\left[3\theta-8\sin\theta-\sin2\theta\right]_{\pi/3}^{5\pi/3}=2\pi+\dfrac{9\sqrt3}{4}\approx10.18$.

EX 2

The other side of the same pair

roles reversed

Find the area inside $r=2+2\cos\theta$ and outside $r=3$.

Use the same intersections

$\theta=\pm\dfrac{\pi}{3}$.

Decide which is outer here

At $\theta=0$: the cardioid gives $4$, the circle gives $3$. The cardioid is outside on $\left(-\dfrac{\pi}{3},\dfrac{\pi}{3}\right)$.

Set up

$A=\dfrac12\displaystyle\int_{-\pi/3}^{\pi/3}\left[\left(2+2\cos\theta\right)^{2}-9\right]d\theta$.

Simplify the bracket

$6+8\cos\theta+2\cos2\theta-9=-3+8\cos\theta+2\cos2\theta$.

Evaluate

$\dfrac12\left[-3\theta+8\sin\theta+\sin2\theta\right]_{-\pi/3}^{\pi/3}=\dfrac{9\sqrt3}{4}-\pi\approx0.755$. Same curves, different range, different outer curve.

EX 3

The pole intersection

invisible to the algebra

Where do $r=\sin\theta$ and $r=\cos\theta$ intersect?

Solve the equation

$\sin\theta=\cos\theta$ gives $\theta=\dfrac{\pi}{4}$, where both equal $\dfrac{\sqrt2}{2}$.

Check the pole separately

$\sin\theta=0$ at $\theta=0$; $\cos\theta=0$ at $\theta=\dfrac{\pi}{2}$.

Note what that means

Both curves pass through the origin, but at different angles.

Explain the blind spot

The equation $r_1=r_2$ compares radii at the same $\theta$, so a shared point reached at different angles never appears.

State the habit

Always solve $r_1=r_2$ and check whether each curve reaches the pole. Both circles here meet at the origin and at $\left(\tfrac{\sqrt2}{2},\tfrac{\pi}{4}\right)$.

EX 4

A region needing two pieces

split at the crossing

Set up the area of the region common to $r=1$ and $r=2\cos\theta$.

Find where they cross

$1=2\cos\theta$ gives $\cos\theta=\dfrac12$, so $\theta=\pm\dfrac{\pi}{3}$.

Identify the inner curve in each range

For $\lvert\theta\rvert\lt\dfrac{\pi}{3}$ the circle $r=1$ is nearer; for $\dfrac{\pi}{3}\lt\lvert\theta\rvert\lt\dfrac{\pi}{2}$ the curve $r=2\cos\theta$ is nearer.

Recognise the principle

The common region is bounded by whichever curve is closer to the pole — the reverse of the outside-minus-inside set-up.

Write both pieces

$A=\dfrac12\displaystyle\int_{-\pi/3}^{\pi/3}1\,d\theta+2\cdot\dfrac12\int_{\pi/3}^{\pi/2}4\cos^{2}\theta\,d\theta$.

Evaluate

$\dfrac{\pi}{3}+\left[\dfrac{\pi}{3}-\dfrac{\sqrt3}{4}\cdot2\right]$, giving $A=\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\approx1.228$.

Watch out

Square each radius, then subtract

$r_{\text{out}}^{2}-r_{\text{in}}^{2}$, never $\left(r_{\text{out}}-r_{\text{in}}\right)^{2}$. The gap between two radii is not itself a radius.

The pole hides from $r_1=r_2$

Two curves can both reach the origin at different angles. Check each curve for $r=0$ as a separate step.

Which curve is outer can change

Test an angle inside each range. When the roles swap, split the integral there and add the pieces.

‘Common to both’ uses the nearer curve

For an intersection region the boundary is whichever curve is closer to the pole in that range — often requiring two pieces.

Course map. Topic 9.9 · LO CHA-5.D — calculate areas of regions defined by polar curves using definite integrals. BC-only.