Area swept in circular sectors rather than rectangular strips, reading the limits from where the distance vanishes, area between two curves and the intersection problem, and arc length in polar form.
Subject: Calculus II · 67 slides · symbolic lesson
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Title
Calculus II · Section 7.4
Sectors instead of strips, and length from the parametric formula
Objectives
Section 7.3 taught you to draw curves such as roses, cardioids and limaçons. This lesson measures them: the area they enclose, the area between two of them, and their length.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 662 — learning objectives 7.4.1 and 7.4.2
In the last section you learned to draw polar curves: roses, cardioids, limaçons, circles through the pole. This section asks the two measuring questions you already know how to answer for ordinary graphs. How much area does the curve enclose, and how long is it?
The answers split neatly. Length turns out to need nothing new at all: a polar curve is a parametric curve with the angle as its parameter, so the Section 7.2 formula does the job after one line of algebra. Area is different. The thin rectangles of a Riemann sum do not fit a region bounded by rays from the pole, so you will rebuild the Riemann sum out of thin circular sectors.
Most of the real difficulty in this section is not the integration, which is mostly trig identities you have seen before. It is choosing the limits so that the region is covered exactly once, and finding every point where two curves cross. Those two skills get whole parts of the lesson.
Warm-up
Discussion prompt
Write down (1) the area under y = f(x) from a to b as a limit of a Riemann sum, and (2) the length of a parametric curve (x(t), y(t)) from t = a to t = b. Then write x and y in terms of r and theta.
Write all three answers down before revealing. Each one comes back later today, and it is much easier to follow a derivation when its ingredients are already on your page.
The Riemann sum is the one to think hardest about. The area formula in this section is built exactly the same way: cut the region into thin pieces, approximate each piece by a shape whose area you know, add, and take a limit. Only the shape changes.
The parametric length formula will be reused almost untouched: you will substitute the conversion equations, x equals r cos theta and y equals r sin theta, into it and watch most of the algebra cancel.
Section
Part 1
Concept
Figure (svg): A dashed circle of radius r centred at the pole, with one sector of central angle theta shaded; the two bounding radii are drawn and a small arc marks the angle theta at the centre.
A full turn is two pi radians, so a sector with central angle theta is the fraction theta over two pi of the whole disc. Take that fraction of the disc's area.
\[ A = \frac{\theta}{2\pi}\cdot \pi r^2 \]
\[ A = \frac12 \theta r^2 \]
The pi cancels and the two survives: that is where the one half in every polar area formula comes from.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 663 — Figure 7.40
Everything in the area half of this section rests on this one small fact, so make sure you could reproduce it with the book closed. A full turn is two pi radians. A sector with central angle theta is therefore theta over two pi of a full turn, and its area is that same fraction of the disc's area, pi r squared.
Multiply it out and the pi cancels, leaving one half theta r squared. Notice where the half comes from: it is the two in two pi, left behind after the pi cancels. It is not a convention, and it is not optional.
Look at the figure. The shaded wedge is bounded by two radii and an arc of the circle. Keep that picture of a wedge in mind; on the next few slides a polar region gets chopped into many thin wedges like it.
Intuition
A rectangle fits the Cartesian grid: a thin vertical strip is exactly what you get between x and x plus dx. The polar grid is made of rays from the pole and circles around it, so the natural thin piece is the wedge between two neighbouring rays.
A polar curve tells you, for each angle, how far out to go. Between two close rays the curve is almost at a constant distance, so the piece of region between those rays is almost a circular sector. That is the whole idea.
Rectangles would be a poor fit here: a polar region is bounded by rays and by a curve given as a distance from the pole, not as a height above an axis.
In Cartesian coordinates you describe a region by heights above the x-axis, so the natural thin piece is a vertical strip, and its area is height times width. A polar curve does not give you heights. It gives you, for each direction, a distance from the pole.
So cut the region along rays from the pole instead. Between two rays that are very close together, the curve is at nearly the same distance the whole way across, which means that thin piece is nearly a circular sector with that distance as its radius.
This is the only genuinely new idea in the area half of the section. Once you accept that a thin polar slice is a sector, the rest is the Riemann sum machinery you already know.
Concept
Figure (svg): The curve r equals one plus sine theta in the first quadrant, from theta equals 0 to pi over 2, with eight thin circular sectors fanned out from the pole; each sector's radius is the curve's value at the sector's far edge, so each one pokes slightly past the curve.
Split the angle interval from alpha to beta into n equal pieces, and give the ith wedge the radius of the curve at its far edge.
\[ \Delta\theta = \frac{\beta - \alpha}{n}, \qquad \theta_i = \alpha + i\,\Delta\theta \]
\[ A_i \approx \frac12\,\Delta\theta\,\big[f(\theta_i)\big]^2 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, pp. 662-663 — Figure 7.39 and the sector area
The figure shows the region in the first quadrant inside the cardioid r equals one plus sin theta, cut into eight equal wedges of angle delta theta. Each wedge is drawn as a true circular sector, with its radius read off the curve at the wedge's far edge.
Because this curve moves away from the pole as theta increases, the far edge is where each slice is widest, so every sector pokes a little past the curve. The sum of the sectors is therefore slightly too big. That is the same thing that happens with a right-endpoint Riemann sum for an increasing function.
The formula for one wedge is just the sector area, one half times the angle times the radius squared, with the radius equal to f at the partition point theta sub i. The book uses exactly these partition points, alpha plus i delta theta.
Concept
Add the sectors, then let the wedges get thinner. The sum is a Riemann sum for the function one half f squared, so its limit is an integral.
\[ A_n = \sum_{i=1}^{n} \frac12\big[f(\theta_i)\big]^2\,\Delta\theta \]
\[ A = \lim_{n\to\infty} A_n = \frac12\int_{\alpha}^{\beta}\big[f(\theta)\big]^2\,d\theta \]
Area of a region bounded by a polar curve (Theorem 7.6) — If f is continuous and nonnegative for theta from alpha to beta, with beta minus alpha between 0 and two pi, the region between the curve r = f(theta) and the rays theta = alpha and theta = beta has the area below.
\[ A = \frac12\int_{\alpha}^{\beta}\big[f(\theta)\big]^2\,d\theta = \frac12\int_{\alpha}^{\beta} r^2\,d\theta \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 663 — Theorem 7.6, equation 7.9
Adding the n sectors gives a sum of terms of the form one half f of theta sub i squared, times delta theta. That is a Riemann sum for the function one half f squared on the interval from alpha to beta, exactly as a rectangle sum is a Riemann sum for f.
As the number of sectors grows, the sectors hug the curve more and more closely, and the Riemann sum becomes the integral. That limit is Theorem 7.6, equation 7.9 in the book.
Read the hypotheses too. The function must be continuous and nonnegative, and the angle interval must be no more than one full turn. The nonnegative condition is there so that the rays and the curve really do bound a region in the simple way the picture shows; you will see shortly how negative values of r are handled.
Notation
Annotate
On: \( A = \frac12\int_{\alpha}^{\beta} r^2\,d\theta \)
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 663 — equation 7.9
Step through the notes one at a time and connect each one to the sector picture. The half and the square are both inherited from one half theta r squared. If you ever doubt the formula, rebuild it from a single sector.
The d theta deserves a moment. The width of each slice is an angle, measured in radians, so the integral is taken with respect to theta. You never convert to x to find a polar area.
The limits are the part of the formula that takes judgment. The formula is only correct when the rays from alpha to beta sweep the region once. Part 2 of this lesson is entirely about choosing them.
Concept
A formula you have just derived should reproduce something you already know. The circle of radius a is the polar curve r equal to a, swept once by a full turn.
\[ A = \frac12\int_0^{2\pi} a^2\,d\theta \]
\[ A = \frac12 a^2 \cdot 2\pi = \pi a^2 \]
Exactly the area of a disc. Exercise 188, the region enclosed by r equal to 4, is this with a equal to 4: sixteen pi.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 188
Any time you derive a formula, test it on a case whose answer you already know. The simplest polar curve is r equal to a constant, a circle about the pole, and a full turn sweeps it once.
The integrand is the constant a squared, so the integral is just a squared times the length of the interval, two pi. The half cancels the two and you get pi a squared, the area of a disc. The formula passes, and so does the factor of one half: without it you would get twice the right answer.
Exercise 188 in the book is this exact computation with a equal to 4. Sixteen pi.
Estimation
\[ A = \frac12\int_0^{\pi/2}(1 + \sin\theta)^2\,d\theta \]
Predict first
With four sectors, radii taken at each wedge's far edge, the sum is 2.460. The radius 1 + sin θ grows on this interval. Is 2.460 too big or too small, and what happens to the error when you double the number of sectors?
Correct: Too big; doubling n roughly halves the error
Why: The radius increases with θ, so the far edge is where each wedge is widest: every sector overshoots the curve and the sum is too big. The overshoots are 0.282, 0.144 and 0.073 for 4, 8 and 16 sectors, halving each time, just like a right-endpoint Riemann sum.
| sectors n | 4 | 8 | 16 | 64 | 256 |
|---|---|---|---|---|---|
| sum | 2.460 | 2.322 | 2.251 | 2.196 | 2.183 |
Commit to an answer before revealing. There are two things to reason about: the direction of the error, and how fast it shrinks.
For the direction, think about which edge of each wedge sets its radius. The radius one plus sin theta grows as theta increases across the first quadrant, and the book's partition points are the far edges of the wedges. So each sector uses the largest radius on its slice and sticks out past the curve. Too big.
For the rate, look at the table. The error drops from about 0.28 to 0.14 to 0.07 as the number of sectors doubles from 4 to 8 to 16. Halving the error each time n doubles is the signature of an endpoint Riemann sum, and it is further evidence that the sector sum really is a Riemann sum in disguise.
Picture it
Figure (svg): Dots showing the sector-sum estimates of the first-quadrant area inside r equals one plus sine theta, for 4, 8, 16, 32, 64 and 128 sectors: 2.460, 2.322, 2.251, 2.215, 2.196 and 2.187, falling toward a dashed line at the exact area 2.178.
Every estimate sits above the dashed line, and each doubling of n moves the dot about halfway down to it. The dashed line is the value the integral will give exactly on the next slide.
Each dot is one sector sum, and the horizontal axis doubles the number of sectors at every step. The dots come down toward the dashed line, and each step closes about half of the remaining gap.
The dashed line is the exact area, three pi over eight plus one, which you are about to compute from Theorem 7.6. Seeing the sums settle on it before you do the integral is the numerical version of the limit in the theorem: the integral is what the sector sums are approaching.
Worked example
Find the area of the region in the first quadrant inside the cardioid.
\[ r = 1 + \sin\theta, \qquad 0 \le \theta \le \frac{\pi}{2} \]
Set up equation 7.9
Why: The first quadrant is swept once as theta runs from 0 to pi over 2, and r stays positive there.
\[ A = \frac12\int_0^{\pi/2}(1 + \sin\theta)^2\,d\theta \]
Expand the square
Why: Multiply out before integrating.
\[ (1 + \sin\theta)^2 = 1 + 2\sin\theta + \sin^2\theta \]
Lower the power of sine
Why: Use the half-angle identity for sine squared.
\[ 1 + 2\sin\theta + \frac{1 - \cos 2\theta}{2} = \frac32 + 2\sin\theta - \frac12\cos 2\theta \]
Find an antiderivative
Why: Term by term.
\[ \int\left(\frac32 + 2\sin\theta - \frac12\cos 2\theta\right)d\theta = \frac{3\theta}{2} - 2\cos\theta - \frac{\sin 2\theta}{4} \]
Evaluate from 0 to pi over 2
Why: Both sine-of-double-angle terms vanish; the cosine term gives 0 at the top and minus 2 at the bottom.
\[ \left(\frac{3\pi}{4} - 0 - 0\right) - (0 - 2 - 0) = \frac{3\pi}{4} + 2 \]
Take half
Why: Do not forget the one half in front.
\[ A = \frac12\left(\frac{3\pi}{4} + 2\right) = \frac{3\pi}{8} + 1 \approx 2.178 \]
Figure (svg): The whole cardioid r equals one plus sine theta drawn dashed, with the part in the first quadrant, from theta equals 0 to pi over 2, shaded and outlined.
Check against the sector sums
Why: The sums were decreasing toward this value from above, as the right-edge radii predicted.
\[ 2.460,\; 2.322,\; 2.251,\; \dots,\; 2.183 \;\searrow\; 2.178 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 190
This is the same region as the sector sums, so you can compare. The first quadrant runs from theta equal to 0 to theta equal to pi over 2, and r is positive the whole way, so those are the limits.
The integration is a pattern you will use in almost every example in this section. Expand the square. Replace the sine squared or cosine squared term with the half-angle identity, so that nothing is squared any more. Then integrate term by term. The only trap is the final one half in front of the integral, which is easy to lose after all that work.
The check compares the exact answer with the numbers from the estimation slide. The sums were all too big and were falling, and they are falling toward 2.178. Two independent routes to the same number is a genuine confirmation.
Trap
A formula remembered without the sector behind it:
\[ A = \int_0^{2\pi} r^2\,d\theta \]
For the circle r equal to a:
\[ A = 2\pi a^2 \quad \text{(WRONG)} \]
A disc of radius a has area pi a squared, not twice that. The half is the sector's own: a sector of angle theta has area one half theta r squared.
\[ A = \frac12\int_0^{2\pi} a^2\,d\theta = \pi a^2 \]
This is the most common single slip with polar area. The formula gets remembered as the integral of r squared, the half falls off, and every answer comes out doubled.
The circle is the fastest way to catch yourself. Put r equal to a into your remembered formula. If you get two pi a squared, you know something is wrong, because a disc has area pi a squared. Better still, remember where the half comes from: a sector of angle theta is theta over two pi of a disc, so its area is one half theta r squared.
Picture it
Figure (svg): Two sectors with the same central angle drawn from the pole: a small one of radius 1 and a large one of radius 2. The large one visibly holds four copies of the small one's area.
Two sectors with the same angle, one with twice the radius of the other. The bigger one is twice as wide AND twice as long, so it holds four times the area. Area in a sector scales with the square of the radius.
\[ \frac12\theta(2r)^2 = 4\cdot\frac12\theta r^2 \]
The figure shows two sectors with the same angle, one with twice the radius. The larger one is not twice the area; it is four times. Doubling the radius doubles the length of the sector along its edges and also doubles its width across the arc, and area multiplies both.
That is the geometric reason the integrand in a polar area is r squared and never r alone. It is the same reason a disc of twice the radius has four times the area. If you ever see an integral of r d theta offered as an area, it fails this simple scaling test.
Section
Part 2
Concept
Figure (svg): The graph of r equals 3 sine 2 theta against theta from 0 to 2 pi, a sine wave crossing zero at 0, pi over 2, pi, 3 pi over 2 and 2 pi. The humps where r is positive are shaded green and those where r is negative are shaded red.
A closed loop of a polar curve leaves the pole and comes back to it. At the pole the distance is zero, so the loops are separated by the zeros of r.
\[ r = 3\sin 2\theta = 0 \iff \theta = 0,\; \frac{\pi}{2},\; \pi,\; \frac{3\pi}{2},\; 2\pi \]
Between two consecutive zeros r keeps one sign, and the curve traces exactly one petal. Graph r against theta first: the humps are the petals.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 664 — Example 7.16 and Figure 7.41
This graph is the most useful sketch in the whole section. It is not the rose itself; it is r plotted against theta as an ordinary function. Every place it crosses zero is a moment when the rose passes through the pole.
A petal is traced as the curve leaves the pole and comes back to it, so each petal corresponds to one hump of this graph, between two consecutive zeros. On the green humps r is positive and the petal points in the direction theta. On the red humps r is negative, and the petal is drawn on the opposite side of the pole.
So the recipe for the limits of one loop is: solve r equal to zero, and take two consecutive solutions. Draw this graph whenever you are unsure which interval traces the piece you want.
Worked example
Find the area of one petal of the rose.
\[ r = 3\sin(2\theta) \]
Find where the first petal starts and ends
Why: r is zero when sin 2 theta is zero; the first two zeros bound the first petal.
\[ 3\sin(2\theta) = 0 \quad\Longrightarrow\quad \theta = 0 \text{ and } \theta = \frac{\pi}{2} \]
Set up equation 7.9
Why: Square the radius and bring out the 9.
\[ A = \frac12\int_0^{\pi/2}\big[3\sin(2\theta)\big]^2 d\theta = \frac92\int_0^{\pi/2}\sin^2(2\theta)\,d\theta \]
Lower the power
Why: The half-angle identity with the angle 2 theta.
\[ \sin^2(2\theta) = \frac{1 - \cos(4\theta)}{2} \]
Integrate
Why: The two halves combine into nine quarters.
\[ A = \frac94\int_0^{\pi/2}\big(1 - \cos 4\theta\big)d\theta = \frac94\left[\theta - \frac{\sin 4\theta}{4}\right]_0^{\pi/2} \]
Evaluate
Why: Sine of 2 pi and sine of 0 are both zero.
\[ A = \frac94\left(\frac{\pi}{2} - \frac{\sin 2\pi}{4}\right) - 0 = \frac{9\pi}{8} \approx 3.534 \]
Figure (svg): The four-petal rose r equals 3 sine 2 theta, with the petal in the first quadrant shaded; that petal is traced as theta goes from 0 to pi over 2.
Check against the whole rose
Why: A full turn traces all four congruent petals once each, so it must give four times the petal.
\[ \frac12\int_0^{2\pi}9\sin^2(2\theta)\,d\theta = \frac92\cdot\pi = \frac{9\pi}{2} = 4\cdot\frac{9\pi}{8} \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, pp. 664-665 — Example 7.16
The limits come straight from the previous slide: r is zero at 0 and at pi over 2, and in between it is positive, so that interval traces the petal in the first quadrant exactly once.
The integral is one half times nine sin squared of two theta. The half-angle identity turns sin squared of two theta into one half minus one half cos four theta, and the rest is routine. Watch the constant: one half from the formula, nine from the square, and another one half from the identity, making nine quarters.
The check uses the symmetry of the rose. This rose has four identical petals, and because two is even, one full turn of theta traces each of them exactly once, including the two drawn with negative r. So the full-turn integral should be four times the petal, and it is: nine pi over two.
Worked example
Find the area inside the cardioid.
\[ r = 1 - \cos\theta \]
Find the zeros of r
Why: The cardioid reaches the pole only where cos theta is 1.
\[ 1 - \cos\theta = 0 \iff \theta = 0,\; 2\pi \quad\Longrightarrow\quad 0 \le \theta \le 2\pi \]
Set up equation 7.9
Why: One full turn sweeps the region once.
\[ A = \frac12\int_0^{2\pi}(1 - \cos\theta)^2\,d\theta \]
Expand and lower the power
Why: Cos squared becomes one half plus one half cos 2 theta.
\[ (1 - \cos\theta)^2 = \frac32 - 2\cos\theta + \frac12\cos 2\theta \]
Integrate
Why: Term by term.
\[ A = \frac12\left[\frac{3\theta}{2} - 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{2\pi} \]
Evaluate
Why: Every sine term is zero at both ends.
\[ A = \frac12\cdot\frac{3\cdot 2\pi}{2} = \frac{3\pi}{2} \approx 4.712 \]
Figure (svg): The cardioid r equals 1 minus cosine theta, with its cusp at the pole and its widest point at x equals minus 2, shaded inside.
Check by scaling
Why: The cardioid r = 2 + 2 cos theta of Example 7.18 is this one reflected and doubled in size, so its area must be four times as big.
\[ \frac12\int_0^{2\pi}(2 + 2\cos\theta)^2 d\theta = \frac12(8\pi + 4\pi) = 6\pi = 4\cdot\frac{3\pi}{2} \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 665 — Checkpoint 7.15
A cardioid has one loop, and it touches the pole only at its cusp. Solving one minus cos theta equal to zero gives theta equal to 0 and to two pi, so a full turn is exactly one tracing, and those are the limits.
The algebra is the same pattern again: expand, replace cos squared with the half-angle identity, integrate. At 0 and two pi every sine term vanishes, so only the three halves theta term survives, and the area is three pi over two.
The check is a scaling argument you can use often. The cardioid in Example 7.18 has the same shape, reflected and with every distance doubled. Doubling every length multiplies areas by four, so its area must be six pi. A quick direct computation confirms it.
Concept
Integrating over a full turn is safe only if a full turn sweeps the region once. For the roses it depends on whether the number in front of theta is even or odd.
| curve | petals | θ-interval that traces it once | one petal |
|---|---|---|---|
| r = 3 sin 2θ (even) | 4 | 0 to 2π | 0 to π/2 |
| r = cos 3θ (odd) | 3 | 0 to π | −π/6 to π/6 |
| r = 1 − cos θ | (one loop) | 0 to 2π | — |
| r = 3 sin θ | (a circle) | 0 to π | — |
For an odd rose, and for circles through the pole, the second half-turn retraces the first with negative r. A full turn counts the region twice.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.3, pp. 652-653 — the petal-count rule, recalled
Here is the fact that makes limits tricky. A full turn of theta, from 0 to two pi, does not always trace a curve exactly once. For the rose with an even multiplier, like three sin two theta, it does. For a rose with an odd multiplier, like cos three theta, the curve is finished after half a turn, and the second half-turn goes round the same petals again with negative r.
The same thing happens for a circle through the pole, like r equals three sin theta: from 0 to pi you go round once, and from pi to two pi you go round again.
Before you integrate over a full turn, ask yourself whether a full turn really traces the region once. If you are not sure, the graph of r against theta, or a quick sketch plotting a few points, settles it.
Worked example
Find the area of one petal of the three-petal rose.
\[ r = \cos 3\theta \]
Find the zeros nearest the petal on the x-axis
Why: Cos 3 theta is zero when 3 theta is plus or minus pi over 2.
\[ 3\theta = \pm\frac{\pi}{2} \quad\Longrightarrow\quad \theta = \pm\frac{\pi}{6} \]
Set up equation 7.9
Why: The petal lies between those two rays.
\[ A = \frac12\int_{-\pi/6}^{\pi/6}\cos^2 3\theta\,d\theta \]
Lower the power
Why: Cos squared of 3 theta, as one half of one plus cos 6 theta.
\[ \cos^2 3\theta = \frac{1 + \cos 6\theta}{2} \]
Integrate and evaluate
Why: Sin 6 theta is zero at both limits.
\[ A = \frac14\left[\theta + \frac{\sin 6\theta}{6}\right]_{-\pi/6}^{\pi/6} = \frac14\left(\frac{\pi}{6} + \frac{\pi}{6}\right) = \frac{\pi}{12} \]
Figure (svg): The three-petal rose r equals cosine 3 theta, with the petal on the positive x-axis shaded between two dashed rays at theta equals minus pi over 6 and pi over 6.
Check against the whole rose
Why: Theta from 0 to pi traces the three congruent petals once each.
\[ \frac12\int_0^{\pi}\cos^2 3\theta\,d\theta = \frac{\pi}{4} = 3\cdot\frac{\pi}{12} \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 192
The petal of this rose that sits on the positive x-axis is centred on theta equal to 0, so look for the zeros of cos three theta nearest to 0. Three theta equal to plus or minus pi over 2 gives theta equal to plus or minus pi over 6.
After that it is the usual pattern, this time with cos squared, which becomes one half plus one half cos six theta. The sine terms vanish at both limits, leaving pi over 12.
The check uses the odd-rose fact from the previous slide. The rose has three identical petals, and theta from 0 to pi traces each of them once. That integral gives pi over 4, which is exactly three times pi over 12.
Error analysis
Asked for the total area enclosed by r = cos 3 theta, someone writes:
Annotate
On: \( A = \frac12\int_0^{2\pi}\cos^2 3\theta\,d\theta = \frac{\pi}{2} \)
Read the line and look for the mistake before opening the notes. The integration is not where it goes wrong: over a full turn, cos squared of three theta averages one half, and the value pi over 2 is correct for that integral.
The mistake is the interpretation. For this odd rose, a full turn traces the three petals twice, so pi over 2 is twice the area of the rose. The correct total, from 0 to pi, is pi over 4.
Nothing in the computation warns you. That is what makes this error dangerous: a wrong interval produces a perfectly clean number. Only the question of how many times the curve is traced catches it.
Sorting
\[ r = \cos 3\theta \]
Sort into buckets
For the three-petal rose r = cos 3θ, what does the integral of ½r² over each θ-interval measure?
Sort each interval before checking. The zeros of cos three theta are spaced pi over 3 apart, at plus or minus pi over 6, pi over 2, five pi over 6, and so on. Any interval between two consecutive zeros gives exactly one petal.
The interval from pi over 6 to pi over 2 is worth a second look. On it, r is negative, so the petal it traces points into the third quadrant, not the first. It is still one complete petal, with the same area as the others.
Any interval of length pi traces the whole rose once, wherever it starts, because the curve repeats after half a turn. A full turn traces it twice. If you remember one thing, remember that last bucket.
Intuition
When r is negative the point is plotted on the opposite side of the pole, but the sector it sweeps is still a sector, with radius equal to the size of r.
\[ \frac12(-r)^2\,\Delta\theta = \frac12 r^2\,\Delta\theta > 0 \]
Squaring throws the sign away. So the area formula needs no special care when r is negative; what the sign changes is WHERE the region is, and therefore which interval of theta draws the loop you want.
Negative values of r look as though they might make trouble for an area formula. They do not. When r is negative, the point is plotted at the same distance on the opposite side of the pole, and the thin slice it sweeps is still a genuine sector, with radius equal to the size of r.
In the formula, the square takes care of this automatically: minus r, squared, is the same as r squared. So every slice contributes a positive area, whatever the sign of r.
What the sign does change is where the slice is. That is why negative r matters for choosing limits, as with the inner loop coming up, but never for the integrand.
Step zero
\[ r = 1 + 2\cos\theta \]
Discussion prompt
This limaçon has a small loop inside a big one. Before any integral: which interval of theta draws the inner loop, and what is special about r there?
Write your answer before revealing. The instinct is to integrate over a full turn; resist it, because a full turn traces both loops.
The inner loop is traced while r is negative. Solve one plus two cos theta equal to zero to find where r changes sign: cos theta equals minus one half, at two pi over 3 and four pi over 3. Between those angles cos theta is less than minus one half, so r is negative, and the points are plotted back across the pole on the right-hand side. That small stretch of theta draws the inner loop.
Once you have the interval, the integrand is still just one half r squared. The negative sign disappears when you square it.
Worked example
Find the area enclosed by the inner loop.
Find the zeros of r
Why: The same cosine value as the previous slide.
\[ 3 + 6\cos\theta = 0 \iff \cos\theta = -\tfrac12 \iff \theta = \tfrac{2\pi}{3},\; \tfrac{4\pi}{3} \]
Set up equation 7.9
Why: r is negative between the zeros: that stretch draws the inner loop.
\[ A = \frac12\int_{2\pi/3}^{4\pi/3}(3 + 6\cos\theta)^2\,d\theta \]
Expand and lower the power
Why: 36 cos squared theta is 18 plus 18 cos 2 theta.
\[ (3 + 6\cos\theta)^2 = 9 + 36\cos\theta + 36\cos^2\theta = 27 + 36\cos\theta + 18\cos 2\theta \]
Integrate
Why: Term by term.
\[ A = \frac12\Big[27\theta + 36\sin\theta + 9\sin 2\theta\Big]_{2\pi/3}^{4\pi/3} \]
Evaluate each term
Why: The sines of 4 pi over 3 and 2 pi over 3 are minus and plus root 3 over 2; the double angles give plus and minus root 3 over 2.
\[ 27\cdot\tfrac{2\pi}{3} = 18\pi, \quad 36(-\sqrt3) = -36\sqrt3, \quad 9\left(\sqrt3\right) = 9\sqrt3 \]
Combine
Why: Half of the total.
\[ A = \frac12\left(18\pi - 27\sqrt3\right) = 9\pi - \frac{27\sqrt3}{2} \approx 4.892 \]
Figure (svg): The limacon r equals 3 plus 6 cosine theta: a large outer loop reaching x equals 9 and a small inner loop between the pole and x equals 3, which is shaded. Dashed rays at theta equals 2 pi over 3 and 4 pi over 3 mark where r is zero.
Check numerically
Why: Simpson's rule with 8 subintervals on the same integral.
\[ \frac12 S_8 \approx 4.894 \quad\text{vs}\quad 9\pi - \tfrac{27\sqrt3}{2} \approx 4.892 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 207
This limaçon is the one from the step-zero slide scaled up by three, so the zeros are at the same angles, two pi over 3 and four pi over 3. Between them r is negative, and that is the inner loop.
Expanding the square gives a cos squared theta term, which becomes eighteen plus eighteen cos two theta. Evaluating is where the care goes. Sine of four pi over 3 is minus root three over two and sine of two pi over 3 is plus root three over two, so their difference is minus root three. For the double angles, sine of eight pi over 3 is plus root three over two and sine of four pi over 3 is minus root three over two, so that difference is plus root three.
The check is numerical. Simpson's rule with just eight subintervals lands within three thousandths of the exact value. When an exact answer involves several signs and square roots, a quick numerical check like this is the fastest way to catch a slip.
Tweak it
Parameter explorer
The curve shows the area swept by r = 1 + c cos θ as θ runs from 0 to x. Slide c past 1, where an inner loop appears. Does the running area ever decrease? What is its value at x = 2π?
\[ \frac12\int_0^{x}(1 + {c}\cos t)^2\,dt \]
\[ \frac12\int_0^{2\pi}(1 + c\cos\theta)^2 d\theta = \pi\left(1 + \frac{c^2}{2}\right) \]
The curve is the area swept so far as theta runs from 0 up to x, for the limaçon r equals one plus c cos theta. Start with c small, then push it past 1, where the inner loop appears.
The running area never goes down, however you set c. That is the square in the integrand at work: every slice adds a positive amount, even while r is negative. At a full turn the total is pi times one plus c squared over 2.
Here is the catch that the slider makes visible. When c is bigger than 1, that full-turn total is not the area of any one region you would draw. It is the area inside the outer loop plus the area of the inner loop again, because the inner loop lies inside the outer loop and is then swept a second time by the negative values of r. To find a region bounded by a limaçon with a loop, split the integral at the zeros of r.
Section
Part 3
Concept
Figure (svg): A region between an outer wavy polar curve and an inner one, swept from one ray to another. One thin slice is highlighted: a sector of the outer radius with a sector of the inner radius removed.
Between two neighbouring rays, the region between two curves is a big sector with a small sector cut out of it. Subtract the two sector areas, then add up and take the limit.
\[ \frac12 r_{\text{out}}^2\,\Delta\theta - \frac12 r_{\text{in}}^2\,\Delta\theta \]
\[ A = \frac12\int_{\alpha}^{\beta}\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)d\theta \]
The limits are now the angles where the two curves cross, because that is where the region begins and ends.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 665 — area between two polar curves
Look at the highlighted slice in the figure. Between two neighbouring rays, the region between the curves is a thin sector reaching out to the outer curve, with a smaller sector reaching only to the inner curve removed from it. Its area is the difference of the two sector areas.
Each sector area has its own radius squared in it, so the difference is one half times the outer radius squared minus the inner radius squared, times the angle. Add up the slices and take the limit, and you get the formula on the slide.
The limits of integration are now usually the angles where the two curves meet, because that is where the region begins and ends. Finding them is the subject of the next few slides, and it has a surprise in it.
Notation
Annotate
On: \( A = \frac12\int_{\alpha}^{\beta}\left(r_{\text{out}}^2 - r_{\text{in}}^2\right)d\theta \)
Each note is one decision you have to make before integrating. The limits come from setting the two radii equal, but that equation can miss points, so the pole always gets a separate look.
To decide which curve is outer, you do not need a careful graph; test one convenient angle between the crossings and see which radius is bigger. And if the outer or the inner curve changes part-way through the region, the formula applies on each piece separately, so split the integral at the change.
The third note is the one that costs the most marks. Square each radius before subtracting. The next trap slide shows how wrong the other order is.
Worked example
The region lies inside the circle and outside the cardioid. Its limits are where they cross.
\[ r = 6\sin\theta \qquad\text{and}\qquad r = 2 + 2\sin\theta \]
Set the two radii equal
Why: A point where both curves are at the same distance at the same angle.
\[ 6\sin\theta = 2 + 2\sin\theta \]
Solve for sine
Why: Collect the sines.
\[ 4\sin\theta = 2 \quad\Longrightarrow\quad \sin\theta = \frac12 \]
Solve for theta
Why: Sine is one half in the first and second quadrants.
\[ \theta = \frac{\pi}{6} \quad\text{and}\quad \theta = \frac{5\pi}{6} \]
Decide which curve is outer
Why: Test an angle in between, pi over 2.
\[ \theta = \tfrac{\pi}{2}: \quad 6\sin\theta = 6 > 4 = 2 + 2\sin\theta \]
Figure (svg): The circle r equals 6 sine theta and the cardioid r equals 2 plus 2 sine theta. They cross at two marked points where r equals 3 and theta is pi over 6 and 5 pi over 6; the crescent inside the circle and outside the cardioid, above those points, is shaded. Both curves also pass through the pole.
Check the crossing point on both curves
Why: At pi over 6 each curve gives r equal to 3: the same point.
\[ 6\cdot\tfrac12 = 3, \qquad 2 + 2\cdot\tfrac12 = 3 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, pp. 665-666 — Example 7.17 and Figure 7.42
The region is a crescent: inside the circle r equals six sin theta, outside the cardioid r equals two plus two sin theta. Before any integral, find where its boundary switches from one curve to the other.
Setting the radii equal is linear in sin theta, and gives sin theta equal to one half. In the range of angles that trace the circle, from 0 to pi, that happens at pi over 6 and five pi over 6. Testing theta equal to pi over 2, the circle is at 6 and the cardioid at 4, so the circle is the outer curve.
The check substitutes the crossing angle into both equations. Both give r equal to 3, so they really are the same point, which the figure shows at the two gold dots. Notice the pink dot at the pole: both curves pass through it too. Hold on to that; Part 4 comes back to it.
Worked example
Set up the area
Why: Circle outer, cardioid inner, between the crossings.
\[ A = \frac12\int_{\pi/6}^{5\pi/6}\left(36\sin^2\theta - (2 + 2\sin\theta)^2\right)d\theta \]
The circle's part
Why: Half of 36 sine squared is 9 times one minus cos 2 theta.
\[ \frac12\int_{\pi/6}^{5\pi/6}36\sin^2\theta\,d\theta = 9\left[\theta - \frac{\sin 2\theta}{2}\right]_{\pi/6}^{5\pi/6} \]
Evaluate it
Why: Sin of 5 pi over 3 is minus root 3 over 2, sin of pi over 3 is plus root 3 over 2.
\[ 9\left(\frac{2\pi}{3} + \frac{\sqrt3}{2}\right) = 6\pi + \frac{9\sqrt3}{2} \]
The cardioid's part
Why: Expand, then lower the power of sine.
\[ \frac12(2 + 2\sin\theta)^2 = 3 + 4\sin\theta - \cos 2\theta \]
Evaluate it
Why: Antiderivative 3 theta minus 4 cos theta minus one half sin 2 theta.
\[ \Big[3\theta - 4\cos\theta - \tfrac12\sin 2\theta\Big]_{\pi/6}^{5\pi/6} = 2\pi + \frac{9\sqrt3}{2} \]
Subtract
Why: The root-three terms cancel.
\[ A = \left(6\pi + \tfrac{9\sqrt3}{2}\right) - \left(2\pi + \tfrac{9\sqrt3}{2}\right) = 4\pi \approx 12.566 \]
Check numerically
Why: Simpson's rule with 8 subintervals on the whole integrand. (The book's text calls the circle r = 3 sin theta in one sentence: a misprint for 6 sin theta.)
\[ \tfrac12 S_8 \approx 12.569 \quad\text{vs}\quad 4\pi \approx 12.566 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 666 — Example 7.17
With the limits and the outer curve settled, the integral splits into the circle's part minus the cardioid's part. Each part is the familiar pattern: expand if needed, use the half-angle identity, integrate term by term.
The evaluation is the fiddly bit, because of the square roots. For the circle, the sine of two theta term at five pi over 6 and at pi over 6 contributes root three over 4 twice. For the cardioid, the cosine term contributes two root three twice and the sine term root three over 4 twice. Both parts end with nine root three over two, and the root-three terms cancel, leaving the clean answer four pi.
The check is Simpson's rule on the whole integrand, which agrees to two decimal places even with only eight subintervals. One more thing: in its solution the book calls the circle r equals three sin theta in one sentence. That is a misprint for six sin theta; the integral it actually computes uses six sin theta.
Worked example
Find the area inside the off-centre circle and outside the circle about the pole.
Find the crossings
Why: Set the radii equal.
\[ 4\cos\theta = 2 \iff \cos\theta = \tfrac12 \iff \theta = \pm\tfrac{\pi}{3} \]
Set up the area
Why: Between the crossings the off-centre circle is outer (at theta = 0 it is 4, against 2).
\[ A = \frac12\int_{-\pi/3}^{\pi/3}\left(16\cos^2\theta - 4\right)d\theta \]
Use the symmetry
Why: The integrand is even, so take twice the integral from 0.
\[ A = \int_0^{\pi/3}\left(16\cos^2\theta - 4\right)d\theta \]
Lower the power
Why: 16 cos squared is 8 plus 8 cos 2 theta.
\[ 16\cos^2\theta - 4 = 4 + 8\cos 2\theta \]
Integrate and evaluate
Why: Sin of 2 pi over 3 is root 3 over 2.
\[ A = \Big[4\theta + 4\sin 2\theta\Big]_0^{\pi/3} = \frac{4\pi}{3} + 2\sqrt3 \approx 7.653 \]
Figure (svg): The circle r equals 4 cosine theta, centred at x equals 2, and the circle r equals 2 centred at the pole. They cross at the points (1, plus or minus root 3). The part of the first circle outside the second, on the right, is shaded.
Check with plane geometry
Why: The disc of radius 2 about (2, 0) has area 4 pi. The part inside r = 2 is a lens of two radius-2 circles whose centres are 2 apart.
\[ 4\pi - \underbrace{\left(\tfrac{8\pi}{3} - 2\sqrt3\right)}_{\text{lens}} = \frac{4\pi}{3} + 2\sqrt3 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 666 — Checkpoint 7.16
The crossings come from four cos theta equal to 2, at plus and minus pi over 3. At theta equal to 0 the off-centre circle is at 4 and the other at 2, so the off-centre circle is outer.
The region is symmetric about the x-axis, and the integrand is an even function of theta, so you may integrate from 0 to pi over 3 and double, which cancels the one half. That saves one evaluation and a chance to drop a sign.
The check comes from plane geometry, and it is worth knowing. The off-centre circle has radius 2 and area four pi. The part of it inside r equal to 2 is a lens made by two circles of radius 2 whose centres are 2 apart, and the standard lens formula gives eight pi over 3 minus two root three. Subtracting gives exactly the integral's answer.
Trap
For the ring inside r = 3 and outside r = 1:
\[ A = \frac12\int_0^{2\pi}(3 - 1)^2\,d\theta \]
\[ A = 4\pi \quad \text{(WRONG)} \]
A ring is a big disc minus a small one: nine pi minus pi. Each radius must be squared on its own, because each sector's area is its own radius squared.
\[ \frac12\int_0^{2\pi}(9 - 1)\,d\theta = 8\pi \]
This mistake comes from treating a polar area like a Cartesian area between curves, where you do subtract the heights first. In polar coordinates the radii are not heights, and each sector's area depends on its own radius squared.
A ring is the quickest test. The ring between radius 1 and radius 3 is a disc of area nine pi with a disc of area pi removed, so its area is eight pi. Subtracting first gives two squared, times pi, which is four pi: only half the truth here, and for the off-centre circle of Checkpoint 7.16 it gives 2.174 instead of 7.653. There is no fixed ratio, so the error cannot be patched afterwards.
Worked example
Find the area inside both circles.
Find the crossings
Why: Set the radii equal.
\[ 6\sin\theta = 3 \iff \sin\theta = \tfrac12 \iff \theta = \tfrac{\pi}{6},\; \tfrac{5\pi}{6} \]
Name the nearer boundary on each stretch
Why: Inside BOTH means inside whichever curve is closer to the pole there.
\[ 0 \to \tfrac{\pi}{6}: \; 6\sin\theta, \qquad \tfrac{\pi}{6} \to \tfrac{5\pi}{6}: \; 3, \qquad \tfrac{5\pi}{6} \to \pi: \; 6\sin\theta \]
Split the integral
Why: The two outer pieces are mirror images.
\[ A = 2\cdot\frac12\int_0^{\pi/6}36\sin^2\theta\,d\theta + \frac12\int_{\pi/6}^{5\pi/6}9\,d\theta \]
The two small pieces
Why: Half-angle identity, then evaluate.
\[ 18\left[\theta - \tfrac12\sin 2\theta\right]_0^{\pi/6} = 18\left(\frac{\pi}{6} - \frac{\sqrt3}{4}\right) = 3\pi - \frac{9\sqrt3}{2} \]
The middle piece
Why: A sector of radius 3 with angle 2 pi over 3.
\[ \frac12\cdot 9\cdot\frac{2\pi}{3} = 3\pi \]
Add
Why: Both parts together.
\[ A = 6\pi - \frac{9\sqrt3}{2} \approx 11.055 \]
Figure (svg): The circle r equals 6 sine theta and the circle r equals 3 overlap in a lens-shaped region, which is shaded. Its lower boundary comes from the circle r equals 6 sine theta near the pole, and its upper arc from the circle r equals 3.
Check with plane geometry
Why: Two radius-3 circles whose centres, (0, 0) and (0, 3), are 3 apart overlap in a lens of area 2R squared times the angle arccos(d over 2R), minus (d over 2) times root(4R squared minus d squared).
\[ 2\cdot 9\cdot\frac{\pi}{3} - \frac32\sqrt{27} = 6\pi - \frac{9\sqrt3}{2} \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 211
A common interior is the set of points inside both curves. Along each ray, that means going out as far as the nearer curve, so the boundary is whichever curve is closer to the pole on each stretch. Here that switches at the crossings.
Near the pole, from 0 to pi over 6, the circle six sin theta is the nearer one. From pi over 6 to five pi over 6 the circle r equals 3 is nearer. Then the first circle takes over again. So the integral splits into three pieces, and the two outer pieces are mirror images of each other.
The middle piece needs no integration at all: it is a sector of a circle of radius 3 with angle two pi over 3. The check again uses the lens formula, since the region is the overlap of two circles of radius 3 whose centres are 3 apart.
Comparison
Comparison matrix
| region | crossings | outer | inner | area |
|---|---|---|---|---|
| inside 6 sin θ, outside 2 + 2 sin θ | π/6, 5π/6 | 6 sin θ | 2 + 2 sin θ | 4π |
| inside 4 cos θ, outside 2 | ±π/3 | 4 cos θ | 2 | 4π/3 + 2√3 |
| inside 3, outside 1 | none | 3 | 1 | 8π |
Fill in each blank before checking. Each row is one of the problems you have just done, and the columns are the decisions you made in each.
The third row has no crossings at all: two circles about the pole never meet, so the ring is swept by a full turn, and its area is one half times nine minus one, times two pi, which is eight pi. The table makes a point worth noticing. Sometimes the limits come from crossings, sometimes from zeros of r, and sometimes simply from one full turn.
Section
Part 4
Concept
In Cartesian coordinates a point has one name. In polar coordinates it has infinitely many: add any whole number of turns to the angle, or flip the sign of r and add half a turn.
\[ (r,\;\theta) = (r,\;\theta + 2k\pi) = (-r,\;\theta + \pi) \]
The pole is the extreme case: r equal to zero with ANY angle names it.
\[ (0,\;\theta) \text{ is the pole for every } \theta \]
Setting two polar equations equal asks for one angle at which both curves give the same r. Two curves can pass through the same point under different names, and then that equation never sees it.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 666 — the discussion after Checkpoint 7.16
This is the root of the intersection problem. A point in the plane has one Cartesian name, but infinitely many polar names. You can add any number of full turns to the angle, or flip the sign of r and add half a turn.
The pole is the most extreme case of all: any angle, paired with r equal to zero, names it.
Now think about what solving r one equal to r two actually does. It looks for a single angle at which both curves give the same r. If the two curves pass through a point under different names, for instance at different angles, or one with positive r and the other with negative r, the equation cannot see that point.
Picture it
Figure (svg): Graphs of r against theta from 0 to 2 pi: r equals 6 sine theta, a sine wave between minus 6 and 6, and r equals 2 plus 2 sine theta, between 0 and 4. They cross at theta equals pi over 6 and 5 pi over 6, where r equals 3. The first is zero at 0, pi and 2 pi; the second is zero only at 3 pi over 2, so no angle makes both zero at once.
Graph each r against theta. The equation 6 sin theta = 2 + 2 sin theta finds the two gold dots. But the circle is at the pole when theta is 0 or pi, and the cardioid only when theta is 3 pi over 2: the pole is on both curves, never at the same angle.
\[ 6\sin\theta = 0 \iff \theta = k\pi, \qquad 2 + 2\sin\theta = 0 \iff \theta = \tfrac{3\pi}{2} + 2k\pi \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 666 — the third intersection point
Here both curves of Example 7.17 are drawn as graphs of r against theta. Where the graphs cross, both curves are at the same distance in the same direction, and those are the two gold dots the algebra found.
Now look at the hollow circles on the theta-axis. The circle six sin theta is at the pole when theta is 0, pi or two pi. The cardioid is at the pole only when theta is three pi over 2. Both curves pass through the pole, so the pole is a genuine meeting point, but they get there at different angles, so the graphs never cross there.
That is the book's third intersection point. The rule it leads to is simple: after solving the equations together, always ask separately whether both curves pass through the pole.
Worked example
Find every intersection, then the area of the region common to both circles.
Solve the equations together
Why: Divide by 3 cos theta.
\[ 3\cos\theta = 3\sin\theta \iff \tan\theta = 1 \iff \theta = \tfrac{\pi}{4} \]
Check the pole separately
Why: Each curve reaches r = 0, at different angles.
\[ 3\cos\theta = 0 \text{ at } \theta = \tfrac{\pi}{2}, \qquad 3\sin\theta = 0 \text{ at } \theta = 0 \]
Use the mirror symmetry about pi over 4
Why: From 0 to pi over 4 the nearer boundary is the sine circle.
\[ A = 2\cdot\frac12\int_0^{\pi/4}9\sin^2\theta\,d\theta = \frac92\Big[\theta - \tfrac12\sin 2\theta\Big]_0^{\pi/4} \]
Evaluate
Why: Sin of pi over 2 is 1.
\[ A = \frac92\left(\frac{\pi}{4} - \frac12\right) = \frac{9\pi}{8} - \frac94 \approx 1.284 \]
Figure (svg): Two circles of diameter 3: r equals 3 cosine theta, centred on the x-axis, and r equals 3 sine theta, centred on the y-axis. They overlap in a lens between the pole and the point (3 over 2, 3 over 2), which is shaded; a dashed ray at theta equals pi over 4 splits the lens into two mirror halves.
Check with plane geometry
Why: Two circles of radius 3 over 2 with centres (3/2, 0) and (0, 3/2), a distance 3 over root 2 apart: the lens formula gives the same number.
\[ 2\cdot\tfrac94\cdot\tfrac{\pi}{4} - \tfrac{3}{2\sqrt2}\cdot\tfrac{3}{\sqrt2} = \tfrac{9\pi}{8} - \tfrac94 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 200
Solving the two equations together gives tan theta equal to 1, so theta equals pi over 4. But each circle passes through the pole, the first at pi over 2 and the second at 0, so the pole is a second meeting point. The figure shows both.
The lens between the pole and the point at pi over 4 is symmetric about the ray at pi over 4. On the lower half, from 0 to pi over 4, the nearer boundary is the sine circle, so the area is twice one half the integral of nine sin squared theta over that interval.
The check uses the lens formula once more. The two circles have radius three halves and centres three over root two apart, and the formula gives exactly nine pi over 8 minus nine quarters, about 1.284.
Worked example
Find every point where the four-petal rose meets the circle of radius one half.
Solve the equations together
Why: Double angles in one full turn of theta.
\[ \cos 2\theta = \tfrac12 \iff 2\theta = \tfrac{\pi}{3},\;\tfrac{5\pi}{3},\;\tfrac{7\pi}{3},\;\tfrac{11\pi}{3} \]
That gives four angles
Why: Halve each.
\[ \theta = \tfrac{\pi}{6},\;\tfrac{5\pi}{6},\;\tfrac{7\pi}{6},\;\tfrac{11\pi}{6} \]
Rename the circle
Why: The same circle is also the curve r = minus one half.
\[ r = \tfrac12 \quad\text{and}\quad r = -\tfrac12 \quad\text{are the same circle} \]
Solve against the other name
Why: Four more angles.
\[ \cos 2\theta = -\tfrac12 \iff \theta = \tfrac{\pi}{3},\;\tfrac{2\pi}{3},\;\tfrac{4\pi}{3},\;\tfrac{5\pi}{3} \]
Convert to positive r
Why: Add half a turn to each angle.
\[ \left(-\tfrac12,\;\tfrac{\pi}{3}\right) = \left(\tfrac12,\;\tfrac{4\pi}{3}\right), \;\dots \]
Figure (svg): The four-petal rose r equals cosine 2 theta and the circle r equals one half. They meet at eight points. Four, at angles pi over 6, 5 pi over 6, 7 pi over 6 and 11 pi over 6, are solid gold dots; the other four, at pi over 3, 2 pi over 3, 4 pi over 3 and 5 pi over 3, are hollow red dots.
Check one of the new points
Why: At theta = pi over 3 the rose gives r = cos(2 pi over 3), which is minus one half: half a unit from the pole, so on the circle.
\[ \cos\tfrac{2\pi}{3} = -\tfrac12, \quad |r| = \tfrac12 \;\checkmark \]
Stewart, Calculus: Early Transcendentals 8e, §10.4 Areas and Lengths in Polar Coordinates §10.4, pp. 669-673 — the intersection example
This example, which Stewart uses, shows that the pole is not the only point the algebra can miss. Solving cos two theta equal to one half over a full turn gives four angles, and four points. But look at the figure: there are eight.
The other four come from the second name of the circle. The curve r equals minus one half is the same circle, because each of its points is half a unit from the pole. Solving cos two theta equal to minus one half gives four more angles, and at those angles the rose is at minus one half, which is on the circle.
The check takes one of the new points, theta equal to pi over 3. The rose gives cos of two pi over 3, which is minus one half, so the point is half a unit from the pole: on the circle. When one curve is a circle about the pole, always solve against both of its names.
Prediction
\[ r = 3\sin\theta \qquad\text{and}\qquad r = 2 - \sin\theta \]
Predict first
Setting the radii equal gives sin θ = 1/2, so θ = π/6 or 5π/6. How many points do the two curves actually share?
Correct: 2
Why: The two solutions give two points, both at distance 3 over 2. There is no third point at the pole this time: 2 minus sin θ is at least 1 for every angle, so the second curve never reaches the pole. Checking the pole is always necessary; it just does not always add a point.
\[ 2 - \sin\theta \ge 1 > 0 \quad\Longrightarrow\quad \text{the pole is not on the second curve} \]
Commit to a number before revealing. You have already been warned to check the pole, so the tempting answer is three.
Check it properly, though. The first curve does pass through the pole, at theta equal to 0. The second, r equals two minus sin theta, never does: sin theta is at most 1, so r is always at least 1. A point shared by the two curves at the pole would need both to reach it, and only one does.
So there are exactly two crossings, both at distance three halves. Checking the pole is a question you must always ask; the answer is sometimes no.
Matching
Match the pairs
Why: Where both curves pass through the pole, the pole is an extra meeting point that the algebra misses. The circle r = 3 never reaches the pole, so pair (b) has just the two solutions. The rose and the small circle hide four points behind the name r = minus one half.
Match each pair before checking, and for each one ask the two questions from this part: what does solving the equations give, and do both curves pass through the pole?
Pairs (a) and (d) both have the pole as an extra point, because in each pair both curves reach it at different angles. Pair (b) does not, since the circle r equals 3 stays three units from the pole. Pair (c) is the rose and the small circle, where the missed points are not at the pole at all but hide behind the negative name of the circle.
Counterexample
Discussion prompt
Someone claims: two polar curves meet exactly where r1(θ) = r2(θ). Find the most extreme counterexample you can: two polar equations whose curves meet everywhere, yet the equation has no solution at all.
Try to find your own example before revealing. The key is to use two different names for the same thing.
The equations r equals 1 and r equals minus 1 describe the same unit circle. Every point of it is shared. Yet setting them equal gives the statement 1 equals minus 1, which has no solution at all. The algebra finds zero intersections of two curves that coincide completely.
This is the counterexample to keep in your head. It proves that solving the equations together can only ever give you some of the intersections, and it tells you why: a sketch is not optional in this section.
Section
Part 5
Concept
Use the conversion equations with r replaced by f of theta. The angle becomes the parameter.
\[ x = f(\theta)\cos\theta, \qquad y = f(\theta)\sin\theta \]
Differentiate each with the product rule:
\[ \frac{dx}{d\theta} = f'(\theta)\cos\theta - f(\theta)\sin\theta \]
\[ \frac{dy}{d\theta} = f'(\theta)\sin\theta + f(\theta)\cos\theta \]
Now the parametric arc length formula from Section 7.2 applies directly, with theta in place of t.
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 667 — arc length in polar curves
Arc length needs no new geometric idea, because length does not depend on how you describe the points. A polar curve is a parametric curve whose parameter is theta: its points are f of theta cos theta, f of theta sin theta.
Differentiating each coordinate uses the product rule, since both are a product of f of theta with a trig function. That gives the two derivatives on the slide, and they are all that the Section 7.2 formula needs.
Exercise 234 in the book asks you to verify the second of these derivatives. It is a one-line product rule, and worth doing once yourself.
Concept
\[ \left(\frac{dx}{d\theta}\right)^2 = f'^2\cos^2\theta - 2ff'\sin\theta\cos\theta + f^2\sin^2\theta \]
\[ \left(\frac{dy}{d\theta}\right)^2 = f'^2\sin^2\theta + 2ff'\sin\theta\cos\theta + f^2\cos^2\theta \]
Add them. The middle terms are equal and opposite, and each remaining pair collects a sine squared plus cosine squared, which is 1.
\[ \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = f'^2\left(\cos^2\theta + \sin^2\theta\right) + f^2\left(\sin^2\theta + \cos^2\theta\right) = f^2 + f'^2 \]
Arc length of a polar curve (Theorem 7.7) — If f has a continuous derivative for theta from alpha to beta, the length of the graph of r = f(theta) over that interval is the integral below.
\[ L = \int_{\alpha}^{\beta}\sqrt{\big[f(\theta)\big]^2 + \big[f'(\theta)\big]^2}\,d\theta = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 667 — Theorem 7.7, equation 7.10
Now square the two derivatives and add them. Each square has three terms. The middle terms, two f f prime sin theta cos theta, appear once with a minus sign and once with a plus sign, so they cancel.
What is left groups into f prime squared times cos squared plus sin squared, and f squared times sin squared plus cos squared. Both brackets are 1. So the whole sum collapses to f squared plus f prime squared: the square of the radius plus the square of its rate of change.
Put that under the square root in the parametric formula and you have Theorem 7.7. It is a satisfying derivation because so much cancels, and it is short enough that you can redo it whenever you forget the formula.
Picture it
Figure (svg): A piece of a spiral with two nearby points on it, at angles theta and theta plus d theta. From the first point, an arc of constant radius r turns through d theta (length r d theta); then a radial segment of length d r reaches the second point. The straight chord between the two points, d s, is the hypotenuse of this small near-right triangle.
The same formula, read off a picture. Turning through a small angle at radius r moves you r times that angle along a circle; meanwhile the curve moves outward by the change in r. The two moves are perpendicular, so Pythagoras gives the step along the curve.
\[ (ds)^2 \approx (dr)^2 + (r\,d\theta)^2 \quad\Longrightarrow\quad ds = \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \]
The algebra on the previous slide has a picture behind it. Take two nearby points on the curve, at angles theta and theta plus d theta. Getting from one to the other can be split into two moves.
First, stay at radius r and turn through the small angle d theta. That moves you along a circular arc of length r d theta, the gold arc. Then move straight outward by the change in r, the orange segment. Those two moves are at right angles, so the straight step d s along the curve is, very nearly, the hypotenuse of a right triangle with those two legs.
Pythagoras then gives d s squared equal to d r squared plus r squared d theta squared. Divide inside the root by d theta squared and you get exactly the integrand of Theorem 7.7.
Notation
Annotate
On: \( L = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \)
Connect each note to the triangle picture. The r squared term is the sideways leg, the movement around a circle. It is never zero unless you are at the pole, which is why a circle about the pole has length even though its r does not change.
The d r d theta squared term is the outward leg. For a circle about the pole it is zero, and the formula reduces to the integral of r d theta, the familiar arc of a circle, radius times angle.
There is no half anywhere in this formula. The half belongs to sectors and to area. And the limits must trace the curve once, for exactly the same reason as with area: a circle through the pole traced twice has twice the length.
Worked example
Find the arc length of the cardioid.
\[ r = 2 + 2\cos\theta \]
Choose the limits
Why: At theta = 0, r = 4, and the cardioid is traced exactly once as theta goes from 0 to 2 pi.
\[ 0 \le \theta \le 2\pi \]
Differentiate
Why: The derivative of cosine is minus sine.
\[ f'(\theta) = -2\sin\theta \]
Square and add
Why: Expand the first square.
\[ f^2 + f'^2 = 4 + 8\cos\theta + 4\cos^2\theta + 4\sin^2\theta \]
Use cos squared plus sin squared
Why: The last two terms make 4.
\[ f^2 + f'^2 = 8 + 8\cos\theta \]
Write the integral
Why: Factor 4 out of the root.
\[ L = \int_0^{2\pi}\sqrt{8 + 8\cos\theta}\,d\theta = 2\int_0^{2\pi}\sqrt{2 + 2\cos\theta}\,d\theta \]
Check at one angle
Why: At theta = pi over 2 the original sum is 4 + 4 = 8, and the simplified form gives 8 + 0.
\[ \big(2 + 0\big)^2 + (-2)^2 = 8 = 8 + 8\cos\tfrac{\pi}{2} \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, pp. 667-668 — Example 7.18
The cardioid starts at r equal to 4 when theta is 0 and returns there after one full turn, never repeating itself on the way, so the limits are 0 to two pi.
The derivative is minus two sin theta. Squaring and adding, the cos squared and sin squared terms again combine into a single constant, leaving eight plus eight cos theta. Factor four out of the root to tidy it up.
The check at a single angle is a good habit when an expression has been simplified: substitute a convenient angle into the unsimplified sum and into the simplified form, and make sure they agree. At pi over 2 both give 8. It will not catch every error, but it catches most dropped terms.
Worked example
Rewrite with a double-angle identity
Why: Cos 2 alpha = 2 cos squared alpha minus 1, with alpha = theta over 2.
\[ 2 + 2\cos\theta = 4\cos^2\frac{\theta}{2} \]
Take the square root
Why: A square root of a square is an absolute value.
\[ \sqrt{4\cos^2\tfrac{\theta}{2}} = 2\left|\cos\tfrac{\theta}{2}\right| \]
Substitute
Why: Carry the 2 outside.
\[ L = 4\int_0^{2\pi}\left|\cos\tfrac{\theta}{2}\right|d\theta \]
Use the symmetry
Why: The two halves are mirror images, and on 0 to pi the cosine of theta over 2 is positive.
\[ L = 8\int_0^{\pi}\cos\tfrac{\theta}{2}\,d\theta \]
Integrate and evaluate
Why: The antiderivative of cos(theta over 2) is 2 sin(theta over 2).
\[ L = 8\Big[2\sin\tfrac{\theta}{2}\Big]_0^{\pi} = 16(1 - 0) = 16 \]
Figure (svg): The cardioid r equals 2 plus 2 cosine theta, with its point farthest right at x equals 4 and its cusp at the pole. The upper half, theta from 0 to pi, is drawn in one colour and the lower half, theta from pi to 2 pi, in another; each half is labelled length 8.
Check numerically
Why: Simpson's rule with 200 subintervals on the original integrand.
\[ \int_0^{2\pi}\sqrt{(2 + 2\cos\theta)^2 + 4\sin^2\theta}\,d\theta \approx 16.000 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 668 — Example 7.18
The root of two plus two cos theta has no obvious antiderivative. The trick is the double-angle identity, read backwards: two plus two cos theta is four cos squared of theta over two. Now the root is easy, but careful: the square root of a square is an absolute value.
The absolute value matters. For theta between pi and two pi, cos of theta over 2 is negative. The figure shows why you can avoid dealing with that: the two halves of the cardioid are mirror images, so the length from 0 to two pi is twice the length from 0 to pi, and on that interval the cosine is positive and the bars can go.
The antiderivative of cos of theta over 2 is two sin of theta over 2, and the answer is 16. Simpson's rule on the original, unsimplified integrand agrees, which checks every step of the simplification at once.
Worked example
Find the total arc length.
\[ r = 3\sin\theta \]
Choose one tracing
Why: This circle through the pole is traced once as theta runs from 0 to pi.
\[ 0 \le \theta \le \pi \]
Differentiate
Why: Derivative of sine.
\[ f'(\theta) = 3\cos\theta \]
Square and add
Why: A Pythagorean identity.
\[ f^2 + f'^2 = 9\sin^2\theta + 9\cos^2\theta = 9 \]
Integrate
Why: The integrand is the constant 3.
\[ L = \int_0^{\pi}\sqrt9\,d\theta = 3\pi \approx 9.425 \]
Figure (svg): Two panels, each showing the circle r equals 3 sine theta of diameter 3 sitting on the pole. Left: traced as theta goes from 0 to pi. Right: traced again, point for point, as theta goes from pi to 2 pi, with negative r.
Check with geometry
Why: The curve is a circle of diameter 3, radius three halves.
\[ 2\pi\cdot\tfrac32 = 3\pi \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 668 — Checkpoint 7.17
The first decision is the interval. This circle passes through the pole, and like every circle of that kind it is traced once as theta goes from 0 to pi. The figure shows that the second half-turn traces it again.
The integrand is remarkably simple: nine sin squared plus nine cos squared is 9, whose root is 3, a constant. So the length is 3 times the length of the interval, three pi.
The check is plain geometry. The curve is a circle of diameter 3, and its circumference is pi times the diameter, three pi. Integrating from 0 to two pi would have given six pi, twice the truth.
Trap
Copying the length formula for y = f(x):
\[ L = \int_{\alpha}^{\beta}\sqrt{1 + \left(\tfrac{dr}{d\theta}\right)^2}\,d\theta \]
For the quarter circle r = 6, theta from 0 to pi over 2:
\[ L = \tfrac{\pi}{2} \quad \text{(WRONG)} \]
The 1 belongs to Cartesian coordinates, where the step dx is a length. In polar the step d theta is an angle, and at radius r it moves you r d theta. The 1 must be r squared.
\[ L = \int_0^{\pi/2}\sqrt{36 + 0}\,d\theta = 3\pi \]
Exercise 218: a quarter of the circumference 12 pi. The wrong formula gives the same length for every circle about the pole, whatever its size.
This error comes from pattern-matching on the formula for the length of y equals f of x, which has a 1 under the root. It looks almost right, which is what makes it dangerous.
In the Cartesian formula the 1 comes from d x, because the step in x is itself a length. In polar coordinates the step in theta is an angle, and turning through it at radius r moves you r d theta. So the 1 has to become r squared. Test any remembered formula on a circle about the pole: the wrong one gives the same length, pi over 2, for a quarter circle of any radius at all, which cannot be right.
Worked example
Find the length of the curve for theta from 0 to 2.
\[ r = e^{3\theta}, \qquad 0 \le \theta \le 2 \]
Differentiate
Why: Chain rule.
\[ f'(\theta) = 3e^{3\theta} \]
Square and add
Why: Both terms are multiples of e to the 6 theta.
\[ f^2 + f'^2 = e^{6\theta} + 9e^{6\theta} = 10e^{6\theta} \]
Take the root
Why: The root of e to the 6 theta is e to the 3 theta.
\[ \sqrt{10e^{6\theta}} = \sqrt{10}\,e^{3\theta} \]
Integrate
Why: An exponential.
\[ L = \sqrt{10}\int_0^{2}e^{3\theta}\,d\theta = \frac{\sqrt{10}}{3}\Big[e^{3\theta}\Big]_0^{2} = \frac{\sqrt{10}}{3}\left(e^{6} - 1\right) \]
Evaluate
Why: e to the 6 is about 403.43.
\[ L \approx 1.0541 \times 402.43 \approx 424.2 \]
Figure (svg): The logarithmic spiral r equals e to the 3 theta for theta from 0 to 2. It starts at the point (1, 0), too close to the pole to see at this scale, and sweeps out through the first quadrant to a point about 403 units from the pole at angle 2 radians, in the second quadrant.
Check by differentiating the antiderivative
Why: It must give back the integrand.
\[ \frac{d}{d\theta}\left[\frac{\sqrt{10}}{3}e^{3\theta}\right] = \sqrt{10}\,e^{3\theta} \;\checkmark \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 219
For the logarithmic spiral the derivative is a multiple of r itself, so f squared plus f prime squared is a multiple of e to the six theta, and its square root is a multiple of e to the three theta. That makes the integral an easy exponential.
The answer, about 424, might look too big until you look at the figure. The spiral's distance from the pole multiplies by e cubed, roughly twenty, every radian. By theta equal to 2 it is over four hundred units out, and nearly all the length is in the final stretch.
The check differentiates the antiderivative and gets the integrand back. For an antiderivative found by an easy rule, that is the quickest complete check there is.
Worked example
Approximate the length of the spiral of Archimedes.
\[ r = 3\theta, \qquad 0 \le \theta \le \frac{\pi}{2} \]
Differentiate
Why: The derivative is the constant 3.
\[ f'(\theta) = 3 \]
Square, add and factor
Why: Take the 9 out of the root.
\[ \sqrt{9\theta^2 + 9} = 3\sqrt{\theta^2 + 1} \]
Write the integral
Why: No elementary shortcut by substitution.
\[ L = 3\int_0^{\pi/2}\sqrt{\theta^2 + 1}\,d\theta \]
Approximate with Simpson's rule
Why: Four subintervals, h = pi over 8.
\[ L \approx 3\cdot\frac{h}{3}\Big[g_0 + 4g_1 + 2g_2 + 4g_3 + g_4\Big], \quad g_k = \sqrt{\theta_k^2 + 1}, \; h = \tfrac{\pi}{8} \]
\[ L \approx 6.2375 \]
Figure (svg): The spiral of Archimedes r equals 3 theta, drawn faintly from theta equals 0 to pi, with the part from 0 to pi over 2 drawn solid: it leaves the pole heading right and curls up to the point (0, 3 pi over 2) on the y-axis.
Check with bounds and the exact value
Why: The integrand runs from 3 up to 3 root(1 + pi squared over 4); a trigonometric substitution gives the exact value.
\[ 4.712 < L < 8.775 \]
\[ L_{\text{exact}} = \tfrac32\Big[\theta\sqrt{\theta^2+1} + \sinh^{-1}\theta\Big]_0^{\pi/2} \approx 6.2376 \]
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 223
The setup is as easy as any in the section: the derivative is 3, and the integrand simplifies to three times the root of theta squared plus 1. But that integral has no quick antiderivative, which is why the book marks this exercise for technology.
Simpson's rule with only four subintervals already gives 6.2375. The step size is pi over 8, and the weights are the usual one, four, two, four, one.
The check does two things. The integrand runs from 3 at the start to about 5.59 at the end, so the length must lie between 3 times pi over 2 and 5.59 times pi over 2; the estimate does. And a trigonometric substitution does give an exact antiderivative, involving the inverse hyperbolic sine, whose value agrees with Simpson to four decimal places.
Intuition
Arc length did not need a new idea: a polar curve IS a parametric curve, and length does not care which coordinates you use to describe the points. Substituting x = f cos theta and y = f sin theta into Section 7.2's formula was enough.
Area was different. Section 7.2's parametric area formula measures area under a curve down to the x-axis, using thin vertical strips. A polar region is bounded by rays from the pole, so it needed a new element, the sector, and a new formula with a half and a square.
\[ dA = \tfrac12 r^2\,d\theta \qquad\text{vs}\qquad ds = \sqrt{r^2 + (dr/d\theta)^2}\,d\theta \]
It is worth seeing why the two halves of this section feel so different. Length is a property of the points of a curve, and it does not care which coordinates you use to name them. So once you notice that a polar curve is a parametric curve, the old formula applies directly.
Area is a property of a region, and the parametric area formula from Section 7.2 measured the region between a curve and the x-axis, sliced by vertical strips. A polar region is bounded by rays from the pole instead. It needed a different slice, the sector, and a new formula with a half and a square in it.
Comparison
Comparison matrix
| question | area | arc length |
|---|---|---|
| the element | ½ r² dθ | √(r² + (dr/dθ)²) dθ |
| where it comes from | a thin sector | the parametric formula (Pythagoras) |
| circle r = a, one turn | πa² | 2πa |
| a factor of one half? | yes | no |
| limits | sweep the region once | trace the curve once |
Fill in the blanks before checking. This table is a good summary of the whole lesson, and each row is a place where the two formulas get confused.
The circle row is the one to use as a test whenever you are unsure of either formula. A circle of radius a, traced once, must give area pi a squared and length two pi a. If a remembered formula gives anything else, it is wrong.
Section
Part 6
Pattern
Figure (svg): A flow diagram. Sketch the curve or curves. For an area: find the limits from the zeros of r or from the crossings, checking the pole separately; name the outer and inner curves; then integrate one half of outer squared minus inner squared. For a length: find one tracing interval, then integrate the square root of r squared plus r prime squared.
This is the order to work in. The sketch comes first, because both formulas are only as good as their limits, and the limits come from understanding the picture. Graphing r against theta is often the quickest way to see where loops start and end.
For area inside a single curve, the limits are consecutive zeros of r. For area between curves, they are the crossings, and the pole is checked separately. Square each radius before you subtract. For length there is no half, and the interval must trace the curve exactly once.
Finish with a check. Plane geometry supplies many: discs, sectors, lenses, circumferences. Symmetry supplies others. And a few steps of Simpson's rule will confirm any exact answer.
Ranking
Put in order
Order the steps for the area inside one polar curve and outside another.
Why: The sketch tells you what region you want and warns you about crossings the algebra will miss. The limits come before the integrand, because the outer curve can only be named between two known crossings.
Put the steps in order before checking. The one most people get wrong is writing the integrand before they know the limits.
You cannot name the outer curve without knowing where the region begins and ends, because the outer curve can change at a crossing. So the crossings come first, including the check of the pole, then the outer and inner curves, then the integral.
Real world
Figure (svg): An elliptical orbit with the sun at the pole, closer to the right end. Two shaded sectors have equal area: a short, wide one near the sun, spanning a large angle, and a long, thin one at the far end, spanning a small angle.
Discussion prompt
A planet sweeps out area at a constant rate: the rate of change of area is one half r squared times the rate of change of theta. In this orbit the far end is 1.5 units from the sun and the near end 0.5. How many times faster does the planet's angle change at the near end?
Kepler found that a planet's line to the sun sweeps out equal areas in equal times. Put the sun at the pole and describe the orbit in polar coordinates, and that becomes a statement about this section's formula: the area swept per unit time, one half r squared times the rate of change of theta, is constant.
The figure shows two sectors of equal area, one near the sun and one far away. Near the sun the radius is small, so a given area needs a wide angle. Far away the radius is three times bigger, and because area goes with the square of the radius, the same area needs an angle nine times smaller.
So the planet's angle turns nine times faster at the near end. For real comets the ratio of distances can be large, and the effect is dramatic: they swing round the sun in weeks and spend decades in the outer part of their orbits.
Check
Check your understanding
Which integral gives the area of one petal of r = 2 cos 2θ?
Answer: A
Why: The petal on the x-axis lies between the consecutive zeros of cos 2θ at minus pi over 4 and pi over 4, and the integrand is one half of r squared. Its value is pi over 2.
The petal of r equals two cos two theta that sits on the x-axis lies between the zeros at minus pi over 4 and pi over 4, and the integrand must be one half r squared.
Each wrong option makes one classic mistake: a full turn that sweeps all four petals, an integral of r instead of one half r squared, or limits that only cover half the petal.
Check
Check your understanding
What is the area inside the circle r = 2 and outside the circle r = 1?
Answer: B
Why: One half the integral from 0 to 2 pi of (4 minus 1) gives 3 pi, the big disc 4 pi minus the small disc pi.
The ring between the circles of radius 1 and 2 is a big disc minus a small disc, four pi minus pi. The formula gives the same thing, one half times four minus one, times two pi.
The distractors are the three traps of this lesson: squaring the difference of the radii, not removing the inner disc, and dropping the half.
Check
Check your understanding
Which integral gives the length of r = θ² for θ from 0 to π?
Answer: A
Why: Here r squared is theta to the fourth and dr/dtheta is 2 theta, whose square is 4 theta squared. The integrand is the root of their sum.
For r equal to theta squared, r squared is theta to the fourth and the derivative is two theta, whose square is four theta squared. The integrand is the square root of their sum.
The most tempting wrong answer copies the Cartesian formula, with a 1 instead of r squared. Remember the triangle: the sideways leg is r d theta, not d theta.
Explain it to yourself
Discussion prompt
In two or three sentences: where does the one half in the polar area formula come from, and why is r squared rather than r? Then say why the arc length formula has neither a half nor a lone r squared.
Write your explanation before revealing it. If you can explain this, you can rebuild both formulas of the section from scratch whenever you need them.
The half and the square both come from one fact, the area of a sector, which is a fraction of a disc. The length formula comes from a different fact, Pythagoras on a tiny triangle whose legs are the outward step and the sideways step, so it has a square root of a sum of squares and no half.
Exit ticket
\[ \text{Exercise 212: inside } r = 1 + \cos\theta \text{ and outside } r = \cos\theta \]
Discussion prompt
Find the area. Careful: how many times does a full turn trace the circle r = cos θ?
OpenStax Calculus Volume 2, §7.4 Area and Arc Length in Polar Coordinates §7.4, p. 669 — Exercise 212
This exercise pulls together the whole of Part 2 and Part 3, and it has a trap built in. The circle r equals cos theta lies entirely inside the cardioid, so the area is simply the cardioid's area minus the circle's area.
The cardioid is traced once by a full turn, giving three pi over 2. But the circle r equals cos theta passes through the pole and is traced once by theta from minus pi over 2 to pi over 2; a full turn would trace it twice. Its area is pi over 4, the area of a disc of radius one half.
So the answer is five pi over 4. If you integrated both curves over a full turn in a single integral, you subtracted the circle twice and got pi. The integral was fine; the limits were not.
Recap
| question | formula | the limits |
|---|---|---|
| area inside r = f(θ) | A = ½ ∫ r² dθ | consecutive zeros of r, or the given rays; sweep once |
| area between two curves | A = ½ ∫ (r²outer − r²inner) dθ | the crossings, plus a separate check of the pole |
| arc length | L = ∫ √(r² + (dr/dθ)²) dθ | one tracing of the curve |
\[ A = \frac12\int_{\alpha}^{\beta} r^2\,d\theta, \qquad L = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta \]
Next, Section 7.5 turns to the conic sections, which have a neat polar form of their own with the focus at the pole: exactly the orbit in the Kepler picture.
Stewart, Calculus: Early Transcendentals 8e, §10.4 Areas and Lengths in Polar Coordinates §10.4, pp. 669-673 — the same material in Stewart
Three formulas, one hazard. The area inside a polar curve is one half the integral of r squared, built from thin sectors. The area between two curves squares each radius before subtracting. The length of a polar curve comes straight from the parametric formula, and has no half.
The hazard in every case is the limits. The region must be swept, or the curve traced, exactly once, so find the zeros of r, find the crossings, and check the pole separately, because solving two polar equations together cannot see points that the curves reach under different names.
The next section studies the conic sections. Their polar equations, with a focus at the pole, are exactly the orbits in Kepler's picture, and this section's area formula is how his second law is written.
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