Area Between Radially Nested Polar Curves

Introduction

When one polar curve lies outside another along every ray in a range of angles, the region between them can be measured by subtracting the inner polar area from the outer polar area. Because polar area depends on the square of the radius, the calculation uses a difference of squared radii, not a difference of radii.

The key condition is that the same curve remains outer throughout the interval. When that condition holds, the region can be described by an outer radius and an inner radius at each angle, and the area has a direct integral formula.

Radial Nesting and the Area Formula

Definition: Let $r_{\text{out}}(\theta)$ and $r_{\text{in}}(\theta)$ be continuous, nonnegative polar radii on an interval $[\alpha,\beta]$. They are radially nested on that interval if $r_{\text{out}}(\theta)\ge r_{\text{in}}(\theta)$ for every $\theta\in[\alpha,\beta]$. The region between them consists, on each ray in the interval, of points whose distance from the pole lies between the two radii.

For the interval to describe the intended region without counting parts more than once, its angles must trace that region once. The nonnegative-radius condition ensures that each radius refers to the ray at angle $\theta$, rather than to the opposite ray through a negative signed radius.

Theorem: If the curves are radially nested on $[\alpha,\beta]$, and that interval traces the region between them once, then its area is $$ A=\frac12\int_{\alpha}^{\beta} \left(r_{\text{out}}(\theta)^2-r_{\text{in}}(\theta)^2\right)\,d\theta. $$

Continuity is a convenient sufficient condition for the radii to be integrable and for the region to have a well-defined area. Equality of the radii at isolated angles, including at the endpoints, is allowed. At such an angle the radial thickness is zero.

Why the Formula Uses Squared Radii

For a fixed angle, the region occupies the radial interval from $r_{\text{in}}(\theta)$ to $r_{\text{out}}(\theta)$. A thin angular slice of width $\Delta\theta$ is approximately an annular sector. A sector of radius $r$ and angle $\Delta\theta$ has area $\frac12r^2\Delta\theta$, so the slice between the two curves has approximate area $\frac12(r_{\text{out}}^2-r_{\text{in}}^2)\Delta\theta$. Adding such slices and taking the limit gives the integral.

The same argument can be stated as a proof using the polar area formula. For a nonnegative polar radius $r(\theta)$ traced once over $[\alpha,\beta]$, the area it encloses is $\frac12\int_\alpha^\beta r(\theta)^2\,d\theta$. Radial nesting means that the inner region is contained in the outer one. Therefore, the area between the curves is the area enclosed by the outer curve minus the area enclosed by the inner curve.

Write each enclosed area.
The outer region has area $\frac12\int_\alpha^\beta r_{\text{out}}(\theta)^2\,d\theta$, and the inner region has area $\frac12\int_\alpha^\beta r_{\text{in}}(\theta)^2\,d\theta$.

Subtract the inner area.
Since the inner region is contained in the outer region, the difference of their areas is the area between the curves.

Combine the integrals.
By linearity of integration, the difference is $\frac12\int_\alpha^\beta(r_{\text{out}}^2-r_{\text{in}}^2)\,d\theta$, as claimed.

This derivation explains why subtracting the two polar areas works: the nesting condition makes one region a subset of the other. It also identifies the limits of the formula. The radii must be ordered consistently on the interval, and the interval must trace the desired region once.

Applying the Formula

A reliable setup begins by naming the outer and inner radii, recording the angular bounds, and checking the inequality between the radii on those bounds. Only then should the squared radii be substituted into the formula. Keeping the factor $\frac12$ outside the integral helps make the structure visible.

Worked Example: Two Concentric Circles

Find the area between the circles $r=3$ and $r=2$, traced over $0\le\theta\le2\pi$.

Both radii are nonnegative, and $3\ge2$ at every angle. Thus $r_{\text{out}}=3$, $r_{\text{in}}=2$, and the interval traces the full annulus once.

\begin{aligned} A &=\frac12\int_0^{2\pi}(3^2-2^2)\,d\theta\\ &=\frac12\int_0^{2\pi}(9-4)\,d\theta\\ &=\frac12\int_0^{2\pi}5\,d\theta\\ &=\frac12[5\theta]_0^{2\pi}\\ &=\frac12(10\pi-0)\\ &=5\pi. \end{aligned}

Check: the outer circle has area $\pi(3)^2=9\pi$, and the inner circle has area $\pi(2)^2=4\pi$. Their difference is $9\pi-4\pi=5\pi$, agreeing with the integral.

Worked Example: Nested Circles with Nonconstant Polar Radii

Find the area between $r=4\cos\theta$ and $r=2\cos\theta$ for $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$.

On this interval, $\cos\theta\ge0$, so both radii are nonnegative. Also $4\cos\theta\ge2\cos\theta$, so the first curve is outer throughout. Substitution gives

\begin{aligned} A &=\frac12\int_{-\pi/2}^{\pi/2} \left((4\cos\theta)^2-(2\cos\theta)^2\right)\,d\theta\\ &=\frac12\int_{-\pi/2}^{\pi/2} (16\cos^2\theta-4\cos^2\theta)\,d\theta\\ &=6\int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta. \end{aligned}

Use $\cos^2\theta=\frac{1+\cos(2\theta)}2$, so an antiderivative is $\frac{\theta}{2}+\frac{\sin(2\theta)}4$. Therefore,

\begin{aligned} A &=6\left[\frac{\theta}{2}+\frac{\sin(2\theta)}4\right]_{-\pi/2}^{\pi/2}\\ &=6\left(\frac{\pi}{4}+0-\left(-\frac{\pi}{4}+0\right)\right)\\ &=6\left(\frac{\pi}{2}\right)\\ &=3\pi. \end{aligned}

Check: $r=4\cos\theta$ describes the circle $x^2+y^2=4x$, with center $(2,0)$ and radius $2$. Similarly, $r=2\cos\theta$ describes the circle $x^2+y^2=2x$, with center $(1,0)$ and radius $1$. The area difference is $4\pi-\pi=3\pi$, matching the integral.

Worked Example: A Noncircular Outer Curve

Find the area between $r=2+\cos\theta$ and $r=1$ for $-\pi\le\theta\le\pi$.

Because $-1\le\cos\theta\le1$, the outer radius satisfies $2+\cos\theta\ge1$. Both radii are nonnegative, and the interval traces the curves once. Thus $r_{\text{out}}=2+\cos\theta$ and $r_{\text{in}}=1$. Expand the difference of squares:

\begin{aligned} (2+\cos\theta)^2-1^2 &=4+4\cos\theta+\cos^2\theta-1\\ &=3+4\cos\theta+\cos^2\theta. \end{aligned}

Using $\cos^2\theta=\frac{1+\cos(2\theta)}2$, an antiderivative of the integrand is $3\theta+4\sin\theta+\frac{\theta}{2}+\frac{\sin(2\theta)}4$. Hence

\begin{aligned} A &=\frac12\int_{-\pi}^{\pi}(3+4\cos\theta+\cos^2\theta)\,d\theta\\ &=\frac12\left[3\theta+4\sin\theta+\frac{\theta}{2} +\frac{\sin(2\theta)}4\right]_{-\pi}^{\pi}\\ &=\frac12\left(\frac{7\pi}{2}-\left(-\frac{7\pi}{2}\right)\right)\\ &=\frac{7\pi}{2}. \end{aligned}

Check the antiderivative by differentiating: $3+4\cos\theta+\frac12+\frac12\cos(2\theta)=3+4\cos\theta+\cos^2\theta$. Its endpoint difference is $7\pi$, and the required factor of $\frac12$ gives $7\pi/2$.

A Common Mistake: Subtracting the Radii

The formula does not use the difference $r_{\text{out}}-r_{\text{in}}$ as the area integrand. That difference describes radial thickness, but area in polar coordinates depends on the square of the radius.

Common mistake: For the annulus with radii $3$ and $2$, an incorrect calculation is $$ \frac12\int_0^{2\pi}(3-2)\,d\theta =\frac12(2\pi)=\pi. $$ The correction is to square each radius: $$ \frac12\int_0^{2\pi}(3^2-2^2)\,d\theta =\frac12(5)(2\pi)=5\pi. $$ The incorrect result also fails the geometric check: subtracting the circle areas gives $9\pi-4\pi=5\pi$, not $\pi$.

Another important check is the ordering. The integrand $r_{\text{out}}^2-r_{\text{in}}^2$ should be nonnegative at every angle under the nesting assumption. If the radii are not ordered consistently, the single formula with that fixed assignment does not describe the area between them over the whole interval.

Why This Method Matters

Many polar regions are naturally described by an inner boundary and an outer boundary along each ray. The method avoids breaking such a region into small geometric pieces: once the angle interval and radial ordering are established, one integral accounts for the entire area. It also reinforces a general feature of area calculations: the relevant quantity is the difference in the areas swept out to the two boundaries, which is why squared radii appear.

Before evaluating, check three things: the radii are nonnegative on the chosen interval, the outer radius is at least the inner radius throughout, and the interval traces the region exactly once. Then insert the radii into the difference-of-squares formula, evaluate the integral, and compare the result with any available geometric or algebraic check.

Check Your Understanding

Use the nesting condition and the polar area formula to answer these questions.

State the conditions needed to use one integral of the squared-radius difference for the area between two curves.
Set up, but do not evaluate, the area between $r=5$ and $r=3$ on $0\le\theta\le\pi$.
For $r_{\text{out}}=3\cos\theta$ and $r_{\text{in}}=\cos\theta$ on $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$, identify which radius is outer and write the area integral.
Explain why the area integrand is based on $r_{\text{out}}^2-r_{\text{in}}^2$, rather than $r_{\text{out}}-r_{\text{in}}$.
What does a negative value of $r_{\text{out}}^2-r_{\text{in}}^2$ at some angle indicate about the claimed outer and inner assignment?