G3 - Areas inside polar curves

G3 - Areas inside polar curves

Polar areas are built from thin sectors rather than vertical strips. You will derive the area formula, choose bounds from the actual tracing, and calculate regions inside or between curves. Careful sketches will help you avoid counting a region twice or subtracting the wrong radii.

Area from thin sectors

How can you measure the overlap of the circles r=2cos⁡θr=2\cos\theta and r=2sin⁡θr=2\sin\theta without finding Cartesian strip heights? Use the rays from their shared pole as the slices.

For a circle of fixed radius RR, a sector with angle Δθ\Delta\theta in radians occupies the fraction Δθ/(2π)\Delta\theta/(2\pi) of the circle. Its area is therefore

Δθ2π×πR2=12R2Δθ\frac{\Delta\theta}{2\pi}\times\pi R^2=\frac12R^2\Delta\theta

For a varying radius r=f(θ)r=f(\theta), a very narrow slice is approximately such a sector. If ff is continuous, the smallest and largest radii on the slice bound its area between two sector areas. As the slice widths tend to zero, those approximations lead to the same integral. This explains both the square of the radius and the factor 1/21/2.

[DIAGRAM: asset_name: Building polar area from sectors; asset_slug: aqa7367-polar-area-sectors; caption: A narrow angular slice has area approaching one half r squared times its angle; description: A shaded radial region is divided conceptually into thin sectors, with one narrow sector highlighted. The radius r and small angle delta theta are labelled. The neighbouring panel gives delta A approximately one half r squared delta theta and explains adding sectors as their angles shrink to zero.]
A narrow angular slice has area approaching one half r squared times its angle

Polar area

For a nonnegative, continuous radius r=f(θ)r=f(\theta), the area swept from the pole to the curve between rays θ=α\theta=\alpha and θ=β\theta=\beta is

A=12∫αβ[f(θ)]2 dθA=\frac12\int_\alpha^\beta [f(\theta)]^2\,\mathrm{d}\theta

provided the chosen sweep covers the required region once, with no repeated interior. Piecewise-continuous boundaries can be split into suitable pieces. Use radians and α<β\alpha<\beta.

This polar-area formula is supplied in the formula booklet, on page 8. Selecting the region and bounds remains your task.

Unlike Cartesian area under a graph, the radius is not a signed height. This formula measures areas of radial sectors. Repeated tracing repeats their areas; squaring an inadmissible negative radius does not make its angle admissible under a nonnegative-radius convention.

Worked example 1: a constant-radius sector

Question. Find the area between the pole and r=2r=2, for 0≤θ≤π/30\leq\theta\leq\pi/3.

Plan. Integrate half the squared radius across the given angular width.

A=12∫0π/34 dθ=[2θ]0π/3=2π3\begin{aligned} A&=\frac12\int_0^{\pi/3}4\,\mathrm{d}\theta\\ &=[2\theta]_0^{\pi/3}=\frac{2\pi}{3} \end{aligned}

Answer. 2π/32\pi/3 square units.

Check. This sector is one sixth of a circle of area 4π4\pi, giving 4π/6=2π/34\pi/6=2\pi/3.

Keep the angle interval and replace the constant radius by 2cos⁡θ2\cos\theta. The sector formula still works, but the changing radius must be squared before integration.

Worked example 2: a changing-radius sector

Question. Find the area between the pole and r=2cos⁡θr=2\cos\theta, for 0≤θ≤π/30\leq\theta\leq\pi/3, with r≥0r\geq0.

Plan. Square the full radius and use cos⁡2θ=(1+cos⁡2θ)/2\cos^2\theta=(1+\cos2\theta)/2.

A=12∫0π/34cos⁡2θ dθ=∫0π/3(1+cos⁡2θ) dθ=[θ+12sin⁡2θ]0π/3=π3+34\begin{aligned} A&=\frac12\int_0^{\pi/3}4\cos^2\theta\,\mathrm{d}\theta\\ &=\int_0^{\pi/3}(1+\cos2\theta)\,\mathrm{d}\theta\\ &=\left[\theta+\frac12\sin2\theta\right]_0^{\pi/3}\\ &=\frac{\pi}{3}+\frac{\sqrt3}{4} \end{aligned}

Answer. π/3+3/4\pi/3+\sqrt3/4 square units.

Check. The curve is the circle of radius 1 centred at (1,0)(1,0). Its upper semicircle has area π/2\pi/2. The omitted cap beyond ray π/3\pi/3 is a sector of central angle π/3\pi/3 minus a triangle, area π/6−3/4\pi/6-\sqrt3/4. Subtracting it gives the same result. The answer is also smaller than Worked example 1, since every radius here is at most 2.

The identity turns a squared cosine into terms whose antiderivatives you know. Keep the chain factor when integrating cos⁡2θ\cos2\theta.

Watch out

Wrong: A=12∫2cos⁡2θ dθA=\tfrac12\int 2\cos^2\theta\,\mathrm{d}\theta when r=2cos⁡θr=2\cos\theta. Squaring the entire radius gives r2=4cos⁡2θr^2=4\cos^2\theta, not 2cos⁡2θ2\cos^2\theta.

Change only the radius scale in a new problem.

Selecting bounds and counting once

The bounds describe a sweep of rays, not horizontal coordinates. Before integrating, identify the region, find its boundary rays, and follow the curve through the proposed interval. An endpoint can be a given ray, a pole crossing found from r=0r=0, or a nonzero intersection found by equating radii.

For r=4cos⁡θr=4\cos\theta, −π/2≤θ≤π/2-\pi/2\leq\theta\leq\pi/2 traces the whole circle once. The smaller interval 0≤θ≤π/20\leq\theta\leq\pi/2 traces its upper half. Integrating over that half and doubling is valid; integrating over the whole circle and then doubling is not.

A petal such as r=3sin⁡2θr=3\sin2\theta, 0≤θ≤π/20\leq\theta\leq\pi/2, starts at the pole, reaches radius 3 at π/4\pi/4, then returns to the pole. Those two zero-radius bounds enclose one petal. Over a larger interval, solve the sign restriction again and count the distinct petals; do not integrate blindly across forbidden angles.

For the cardioid r=2(1+cos⁡θ)r=2(1+\cos\theta), the radius stays nonnegative. A sweep 0≤θ≤2π0\leq\theta\leq2\pi traces its complete boundary once, returning through a cusp at π\pi. Although two parameter values label the starting boundary point, that shared endpoint has zero area; it does not duplicate an interior region.

Worked example 3: the complete cardioid

Question. Find the total area enclosed by r=2(1+cos⁡θ)r=2(1+\cos\theta), with r≥0r\geq0.

Plan. Use one full tracing from 0 to 2π2\pi, then expand the square and replace cos⁡2θ\cos^2\theta.

A=12∫02π4(1+2cos⁡θ+cos⁡2θ) dθ=∫02π(3+4cos⁡θ+cos⁡2θ) dθ=[3θ+4sin⁡θ+12sin⁡2θ]02π=6π\begin{aligned} A&=\frac12\int_0^{2\pi}4(1+2\cos\theta+\cos^2\theta)\,\mathrm{d}\theta\\ &=\int_0^{2\pi}(3+4\cos\theta+\cos2\theta)\,\mathrm{d}\theta\\ &=\left[3\theta+4\sin\theta+\frac12\sin2\theta\right]_0^{2\pi}\\ &=6\pi \end{aligned}

Answer. 6π6\pi square units.

Check. Over a full turn, the average values of cos⁡θ\cos\theta and cos⁡2θ\cos2\theta are zero. Thus the expanded integrand has mean 3 over width 2π2\pi, confirming 6π6\pi. The region fits inside the radius-4 circle of area 16π16\pi.

Keep the curve and change only the requested region to its upper half. Symmetry should halve the result, rather than require a new formula.

Worked example 4: half of the same cardioid

Question. Find the area enclosed by r=2(1+cos⁡θ)r=2(1+\cos\theta) above the initial line.

Plan. Trace the upper arc with 0≤θ≤π0\leq\theta\leq\pi and integrate once.

Aupper=12∫0π[2(1+cos⁡θ)]2 dθ=[3θ+4sin⁡θ+12sin⁡2θ]0π=3π\begin{aligned} A_{\text{upper}}&=\frac12\int_0^\pi[2(1+\cos\theta)]^2\,\mathrm{d}\theta\\ &=\left[3\theta+4\sin\theta+\frac12\sin2\theta\right]_0^\pi\\ &=3\pi \end{aligned}

Answer. 3π3\pi square units.

Check. Reflection in the initial line preserves the cardioid, so its upper and lower areas are equal. Twice this result agrees with the independently computed total in Worked example 3.

If signed radii are explicitly allowed, r=4cos⁡θr=4\cos\theta over 0≤θ≤2π0\leq\theta\leq2\pi traces its circle twice: the negative-radius half-turn repeats the positive-radius points. Its integral then gives twice the circle's area. Under the nonnegative-radius convention those negative-radius angles are excluded instead. Either way, tracing must be settled before interpreting the integral as a geometric area.

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