Area in Polar Coordinates

Polar area uses sectors rather than rectangles. A thin sector has area approximately one half of squared radius times its angle. Integration adds those pieces, but the result represents the intended region only when the bounds and tracing are correct.

Petals, negative radii and overlapping curves make setup more important than the antiderivative. Identify which rays belong to the region, which boundary is outermost and how many times each part is covered.

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Derive the sector formula

A sector of radius $r$ and angle $\Delta\theta$ in radians has area $r^2\Delta\theta/2$ . For a nonnegative radial boundary describing a region once,

A=\frac12\int_\alpha^\beta r(\theta)^2\,d\theta.

The angle must be in radians. More generally, polar coordinates have area element $\rho\,d\rho\,d\theta$ ; integrating from $\rho=0$ to the boundary gives the same formula. This interpretation makes the radial interval explicit.

Calculate a cardioid area

For $r=1+\cos\theta$ , the radius is nonnegative and $[0,2\pi]$ traces the cardioid once. Therefore

A=\frac12\int_0^{2\pi}(1+\cos\theta)^2\,d\theta.

Expansion and the identity $\cos^2\theta=(1+\cos2\theta)/2$ give contributions $2\pi$ , zero and $\pi$ inside the brackets. Thus $A=3\pi/2$ .

The cusp is an isolated boundary point and adds no separate area. The interval works because it gives one tracing, not simply because $2\pi$ is a familiar full angle.

Isolate a rose petal

For $r=2\cos3\theta$ , the petal centred on the positive horizontal axis lies between consecutive zeros $-\pi/6$ and $\pi/6$ . Radius is nonnegative there:

A_{\mathrm{petal}}=\frac12\int_{-\pi/6}^{\pi/6}4\cos^2(3\theta)\,d\theta.

Since $2\cos^2(3\theta)=1+\cos6\theta$ ,

A_{\mathrm{petal}}=\left[\theta+\frac{\sin6\theta}{6}\right]_{-\pi/6}^{\pi/6}=\frac\pi3.

The three congruent petals have total area $\pi$ . Integrating the squared-radius expression over $[0,2\pi]$ gives $2\pi$ , because that interval traces the rose twice.

Avoid double-counting the rose

The right petal uses −π/6≤θ≤π/6.

Diagram explanation and exact data

The full curve is drawn on an interval of length π. The squared-radius integral over twice that interval counts every petal twice.

Avoid double-counting the rose: exact reference data
Region	Area
one petal	π/3
three distinct petals	π
0 to 2π tracing integral	2π

The square removes the radius sign from the integrand, not repeated coverage from the geometry. A positive integral can still overcount area.

Subtract squared radial boundaries

For a region described once by

\alpha\le\theta\le\beta,\qquad0\le r_{\mathrm{in}}(\theta)\le\rho\le r_{\mathrm{out}}(\theta),

use

A=\frac12\int_\alpha^\beta(r_{\mathrm{out}}^2-r_{\mathrm{in}}^2)\,d\theta.

It is not the integral of half the squared radial thickness. Subtracting sector areas requires a difference of squares.

For the first-quadrant region between $r=1$ and $r=2$ , the area is $(1/2)(4-1)(\pi/2)=3\pi/4$ . The polar area reference develops this method and its geometric interpretation.

Use the smaller radius for an overlap

The disks bounded by $r=2\sin\theta$ and $r=2\cos\theta$ are unit disks centred at $(0,1)$ and $(1,0)$ . Their overlap is in the first quadrant.

A point on a first-quadrant ray belongs to both disks only up to the smaller radius. The radii agree at $\theta=\pi/4$ . Before that, $2\sin\theta$ is smaller; afterwards, $2\cos\theta$ is smaller. Therefore

A=\frac12\int_0^{\pi/4}(2\sin\theta)^2\,d\theta+\frac12\int_{\pi/4}^{\pi/2}(2\cos\theta)^2\,d\theta.

Each term equals $\pi/4-1/2$ , giving $A=\pi/2-1$ . Choosing the larger radius describes a union of radial portions, not the overlap. Choosing one boundary throughout includes points outside the other disk.

The common region of two disks

Equal axis units; each circle has radius one.

Diagram explanation and exact data

The radial upper bound is 2 sin θ before π/4 and 2 cos θ afterwards. The origin is another boundary intersection.

The common region of two disks: exact reference data
Angle interval	Upper radius	Area
0 to π/4	2 sin θ	π/4−1/2
π/4 to π/2	2 cos θ	π/4−1/2

Check intersections geometrically

The same point can have different polar descriptions, particularly at the pole or with a negative radius. Equal radii at equal angles need not reveal every intersection.

For the two disks, the nonzero common-radius equation identifies $(1,1)$ , while the pole is another boundary intersection. It appears at endpoints of the overlap description rather than through that nonzero common-radius equation.

Convert representative points or an entire curve to Cartesian form when needed. The purpose is to establish the region, not to force every stage of reasoning to remain in polar notation.

Separate geometric area from an oriented integral

For a piecewise-smooth simple closed boundary traced once with positive orientation,

A=\frac12\oint(x\,dy-y\,dx).

Substitution of $x=r\cos\theta$ , $y=r\sin\theta$ produces the expression involving $r^2d\theta$ . Retracing counts multiple coverage; self-intersections can create signed contributions instead of the sum of every visible loop’s area.

This explains the limitation of the formula. Reversing an oriented boundary changes the integral’s sign, while geometric area remains nonnegative. Splitting into simply described petals or radial regions is often the clearest introductory approach.

Decide whether one radial interval is sufficient

Some regions have separated radial portions on the same ray. Others are not bounded by a single nonnegative radial function over the chosen angular interval. Split such regions or use multiple radial bounds in a double integral.

A reliable setup records angular bounds, radial bounds, changes of boundary order and any symmetry multiplier. Symmetry reduces work only after the pieces are shown to be congruent and nonoverlapping except on boundaries.

Practice questions

Find the area of a radius-four sector with angle $\pi/3$ .
Find the first-quadrant area between $r=1$ and $r=3$ .
Why does the interval $[0,2\pi]$ overcount the three-petal rose?

Check your answers

$(1/2)(16)(\pi/3)=8\pi/3$ .
$(1/2)(9-1)(\pi/2)=2\pi$ .
It traces the same three petals twice. One petal has area $\pi/3$ , so the geometric total is $\pi$ , not the repeated-tracing integral $2\pi$ .

Sources and further reading

OpenStax, Calculus Volume 2: Area and Arc Length in Polar Coordinates.