PP3 - Areas enclosed by polar curves

PP3 - Areas enclosed by polar curves

Find exact areas by adding narrow polar sectors. You will choose limits that trace the required region once, handle a given expression for r2r^2, and combine curves by selecting the correct inner and outer radii.

Add sectors and count each region once

What is the area of the lower petal of r=4cos⁡2θr=4\cos2\theta, even though that petal has r<0r<0? Its radial coordinate is signed, but its area is positive. The square in the area formula handles the sign; your limits must handle which petal is counted.

You already know

For a circle, a sector of radius RR and angle Δθ\Delta\theta radians has area 12R2Δθ\frac12R^2\Delta\theta. A polar curve may have negative rr, when its points lie opposite the indicated direction and its distance from the pole is ∣r∣|r|.

Split an angular interval into narrow pieces. In a sufficiently narrow piece, the radius of a continuous curve changes very little, so the area is approximately 12r2Δθ\frac12r^2\Delta\theta. Adding the sectors and taking the limit gives an integral. For negative rr, the distance is ∣r∣|r|, whose square is still r2r^2.

[DIAGRAM: asset_name: Polar area adds narrow sectors; asset_slug: b042_043_area_sectors; caption: Each narrow piece contributes approximately one half r squared times its angle; description: A positive-radius polar curve is drawn between two rays labelled alpha and beta. Its swept region is shaded and one narrow sector is highlighted. That sector has radius r and angular width delta theta, with approximate area one half r squared delta theta. A separate annotation states that the chosen angular interval must sweep the required region once.]
Each narrow piece contributes approximately one half r squared times its angle

Area enclosed by a polar curve and two rays

If a continuous polar arc r=f(θ)r=f(\theta), from θ=α\theta=\alpha to θ=β\theta=\beta, sweeps the required region once without overlap, its area with the two bounding rays is

A=12∫αβr2 dθ,α<βA=\frac12\int_{\alpha}^{\beta}r^2\,d\theta,\qquad \alpha<\beta

For a loop that starts and finishes at the pole, use those pole angles as limits. A closed curve surrounding the pole may instead require a full turn; always check that it is traced once. Negative values of rr are allowed, but identify their actual positions before choosing the limits.

In the formula booklet.

The formula counts every sweep. Here is a useful trap: r=4cos⁡θr=4\cos\theta describes a circle of radius 22, so its area is 4π4\pi. Integrating over 0≤θ≤2π0\le\theta\le2\pi gives

12∫02π16cos⁡2θ dθ=8π\frac12\int_0^{2\pi}16\cos^2\theta\,d\theta=8\pi

because that interval traces the circle twice. The single-pass interval −π/2≤θ≤π/2-\pi/2\le\theta\le\pi/2 gives 4π4\pi, as it should. A correct integrand cannot repair incorrect limits.

Key idea

Find the region on a sketch before integrating. The formula needs r2r^2, the appropriate angular interval and any justified symmetry factor; it does not automatically know which enclosed region you meant.

Sometimes the question gives r2r^2 directly. Substitute that expression once, rather than square it again.

Use a given expression for r squared

An equation for r2r^2 is especially convenient for area: there is no need to take a square root. First find the allowed interval and the zeros at the pole. For a non-negative radial branch, these determine the visible loop.

Worked example 1: a loop with r squared supplied

Question. For r≥0r\ge0, a curve has equation r2=8sin⁡2θr^2=8\sin2\theta. Determine the area of its loop in the first quadrant.

Plan. Use the zeros 00 and π/2\pi/2 as limits, substitute r2r^2 directly, and integrate.

A=12∫0π/28sin⁡2θ dθone first-quadrant loop=4[−12cos⁡2θ]0π/2account for the factor 2=−2(cos⁡π−cos⁡0)=4substitute both limits\begin{aligned} A&=\frac12\int_0^{\pi/2}8\sin2\theta\,d\theta &&\text{one first-quadrant loop} \\ &=4\left[-\frac12\cos2\theta\right]_0^{\pi/2} &&\text{account for the factor }2 \\ &=-2(\cos\pi-\cos0)=4 &&\text{substitute both limits} \end{aligned}

Why this step? The equation already gives the square required by the formula. Taking a square root and then squaring it would add work without changing the integrand.

Answer. The area is 44 square units.

Check. The largest radius is 8=22\sqrt8=2\sqrt2. A quarter-circle sector with that constant radius would have area 2π2\pi, so the smaller value 44 is reasonable.

Change only sine to cosine. The loop turns to lie around the initial line, so its zero-to-zero interval changes.

Worked example 2: a loop centred on the initial line

Question. For r≥0r\ge0, a curve has equation r2=8cos⁡2θr^2=8\cos2\theta. Determine the area of its right-hand loop.

Plan. Find the consecutive zeros on either side of 00, then integrate r2r^2 over that interval.

r=0⟺θ=−π4,π4bounds of the right loopA=12∫−π/4π/48cos⁡2θ dθone sweep=2[sin⁡2θ]−π/4π/4=2(1−(−1))=4integrate and evaluate\begin{aligned} r=0&\Longleftrightarrow\theta=-\frac{\pi}{4},\frac{\pi}{4} &&\text{bounds of the right loop} \\ A&=\frac12\int_{-\pi/4}^{\pi/4}8\cos2\theta\,d\theta &&\text{one sweep} \\ &=2[\sin2\theta]_{-\pi/4}^{\pi/4}=2(1-(-1))=4 &&\text{integrate and evaluate} \end{aligned}

Why this step? cos⁡2θ≥0\cos2\theta\ge0 throughout this interval. The next interval would give a negative r2r^2, rather than a continuation of this real branch.

Answer. The area is 44 square units.

Check. Rotating the first loop through −π/4-\pi/4 preserves its area, so the two answers agree.

The maximum of r2r^2 is not the maximum of rr. Taking that final square root matters when checking the curve's size.

Watch out

Wrong: from r2=8sin⁡2θr^2=8\sin2\theta, use 64sin⁡22θ64\sin^22\theta in the area integral, or say the maximum radius is 88. Right: use 8sin⁡2θ8\sin2\theta directly, and the maximum non-negative radius is 8\sqrt8.

Scale the second curve's squared radius for the next check.

Expand the square before integrating

When rr, rather than r2r^2, is given, square the whole expression. For a curve such as r=a(1+cos⁡θ)r=a(1+\cos\theta), the cross term matters, and a double-angle identity removes the squared cosine.

Reducing a squared trigonometric function

For every real uu,

cos⁡2u=1+cos⁡2u2,sin⁡2u=1−cos⁡2u2\cos^2u=\frac{1+\cos2u}{2},\qquad \sin^2u=\frac{1-\cos2u}{2}

These follow from the compound-angle identities in the formula booklet and sin⁡2u+cos⁡2u=1\sin^2u+\cos^2u=1.

Not in the formula booklet: learn it.

The integration is ordinary trigonometric integration once the square is expanded. The region decides the limits; the formula does not require a whole loop if two rays bound the requested region.

Worked example 3: part of a cardioid

Question. Find the area bounded by r=2(1+cos⁡θ)r=2(1+\cos\theta) and the rays θ=0\theta=0, θ=π/2\theta=\pi/2.

Plan. Square the radius, expand, replace cos⁡2θ\cos^2\theta by a double-angle expression, and integrate between the rays.

A=2∫0π/2(1+2cos⁡θ+cos⁡2θ) dθinclude the cross term=2∫0π/2(32+2cos⁡θ+12cos⁡2θ)dθreduce the square=2[32θ+2sin⁡θ+14sin⁡2θ]0π/2integrate each term=2(3π4+2)=3π2+4evaluate\begin{aligned} A&=2\int_0^{\pi/2}(1+2\cos\theta+\cos^2\theta)\,d\theta &&\text{include the cross term} \\ &=2\int_0^{\pi/2}\left(\frac32+2\cos\theta+\frac12\cos2\theta\right)d\theta &&\text{reduce the square} \\ &=2\left[\frac32\theta+2\sin\theta+\frac14\sin2\theta\right]_0^{\pi/2} &&\text{integrate each term} \\ &=2\left(\frac{3\pi}{4}+2\right)=\frac{3\pi}{2}+4 &&\text{evaluate} \end{aligned}

Why this step? (1+cos⁡θ)2(1+\cos\theta)^2 contains 2cos⁡θ2\cos\theta; omitting it would lose the 44 in the answer.

Answer. 3π2+4\frac{3\pi}{2}+4 square units.

Check. On this interval 2≤r≤42\le r\le4, so the area lies between the sector areas π\pi and 4π4\pi, as the answer does.

Change only the final ray to θ=π\theta=\pi. This asks for the whole upper half of the same cardioid.

Worked example 4: a larger angular region

Question. Find the area bounded by r=2(1+cos⁡θ)r=2(1+\cos\theta) and the initial line in the upper half-plane.

Plan. Use the same antiderivative, now from 00 to π\pi, where the curve reaches the pole.

A=12∫0π4(1+cos⁡θ)2 dθupper half of the curve=2[32θ+2sin⁡θ+14sin⁡2θ]0πsame expanded integrand=2(3π2)=3πboth sine terms vanish\begin{aligned} A&=\frac12\int_0^{\pi}4(1+\cos\theta)^2\,d\theta &&\text{upper half of the curve} \\ &=2\left[\frac32\theta+2\sin\theta+\frac14\sin2\theta\right]_0^{\pi} &&\text{same expanded integrand} \\ &=2\left(\frac{3\pi}{2}\right)=3\pi &&\text{both sine terms vanish} \end{aligned}

Why this step? The interval 0≤θ≤π0\le\theta\le\pi includes the small part of the upper cardioid to the left of the pole; 0≤θ≤π/20\le\theta\le\pi/2 does not.

Answer. 3π3\pi square units.

Check. The extra area is 3π−(3π/2+4)=3π/2−4>03\pi-(3\pi/2+4)=3\pi/2-4>0. Reflection in the initial line gives whole-cardioid area 6π6\pi.

The sketches show why these two limits describe different regions of the same curve.

[DIAGRAM: asset_name: Rays determine which cardioid area is counted; asset_slug: b042_043_cardioid_areas; caption: Ending at pi over 2 gives the first-quadrant part; ending at pi gives the whole upper half; description: Two panels show r equals 2 times (1 plus cosine theta). The first shades the region swept from zero to pi over 2 and labels its area 3 pi over 2 plus 4. The second shades the upper half swept from zero to pi and labels area 3 pi. Each includes the pole, initial line and the relevant ending ray.]
Ending at pi over 2 gives the first-quadrant part

Exam tip

Under detailed reasoning, show the area integral, the square expanded, the trigonometric reduction and the antiderivative before substituting the limits. A calculator's area value alone does not supply that chain of reasoning.

The next curve scales every length in Worked example 3 by 3/23/2. Its area must therefore scale by 9/49/4.

Watch out

Wrong: (1+cos⁡θ)2=1+cos⁡2θ(1+\cos\theta)^2=1+\cos^2\theta. This loses the cross term. The correct expansion is 1+2cos⁡θ+cos⁡2θ1+2\cos\theta+\cos^2\theta; in Worked example 3 that middle term contributes 44 to the area.

Expand the entire squared radius before reducing the squared cosine.

Continue this lesson.

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