Chapter 10 · Parametric Equations and Polar Coordinates
10.4 Area and Arc Length in Polar Coordinates
Slice a polar region into thin pizza slices: $A=\tfrac12\int r^2\,d\theta$ for area and $L=\int\sqrt{r^2+(dr/d\theta)^2}\,d\theta$ for length.
How much ground does a rotating radar actually cover when hills block part of its view? How much sea does a lighthouse beam sweep, how much grass does a rotating sprinkler water, and how long is the edge of a cardioid-shaped cam in an engine? Each of these regions is organised around a centre, so it is described most naturally in polar coordinates — and the vertical rectangles we used for areas in Chapter 6 fit such regions very badly.
In this section we replace rectangles by thin circular sectors (pizza slices) radiating from the pole. Adding up sectors and taking a limit gives the polar area formula $A=\tfrac12\int r^2\,d\theta$. We then find areas of petals and loops, areas between two polar curves (where finding all the intersection points needs real care), and finally the length of a polar curve.
Area of a Region Bounded by a Polar Curve
Recall that a circular sector of radius $r$ and angle $\Delta\theta$ (in radians) is the fraction $\dfrac{\Delta\theta}{2\pi}$ of a whole disk, so its area is
$$\frac{\Delta\theta}{2\pi}\cdot\pi r^2=\tfrac12r^2\,\Delta\theta.$$
Now let $R$ be the region bounded by the curve $r=f(\theta)$ and the rays $\theta=\alpha$ and $\theta=\beta$. Divide $[\alpha,\beta]$ into $n$ subintervals of width $\Delta\theta=\dfrac{\beta-\alpha}{n}$ and pick a sample angle $\theta_i^*$ in the $i$-th one. The thin slice of $R$ between $\theta_{i-1}$ and $\theta_i$ is close to a sector of radius $f(\theta_i^*)$, so
$$A\approx\sum_{i=1}^{n}\tfrac12\left[f(\theta_i^*)\right]^2\Delta\theta.$$
This is a Riemann sum for $\tfrac12f(\theta)^2$; as $n\to\infty$ the approximation becomes exact (Interactive 1 shows the sectors squeezing the true area from inside and outside).
📐 Theorem 10.8 — Area in polar coordinates
Let $f$ be continuous and $f(\theta)\ge0$ on $[\alpha,\beta]$, where $0\lt\beta-\alpha\le2\pi$. The area of the region bounded by the curve $r=f(\theta)$ and the rays $\theta=\alpha$ and $\theta=\beta$ is
Remark. If instead $f(\theta)\le0$ on $[\alpha,\beta]$, the curve is drawn on the opposite rays and the slices are sectors of radius $|f(\theta)|$; since $|f|^2=f^2$ the same formula still gives the area swept out. What matters is that, as $\theta$ runs from $\alpha$ to $\beta$, the radius arm sweeps over each point of the region exactly once.
Interactive 1 — Pizza-Slice Approximation of a Polar Area
The region is cut into $n$ equal angles. On each one the green sector uses the smallest value of $|r|$ (it fits inside the region) and the blue sector the largest (it covers the slice). The true area $\tfrac12\int r^2\,d\theta$ is trapped between the two sums, and the gap closes as $n$ grows. Press Play to double the slices again and again.
slices $n$, angle $\Delta\theta$—
inner sum (green)—
outer sum (blue)—
exact area—
gap (outer − inner)—
🧭 Problem-solving strategy — finding a polar area
Sketch the curve (Section 10.3) and shade the region you want.
Find the limits. Determine the interval of $\theta$ over which the radius arm sweeps the region exactly once. For a petal or loop this is between two consecutive zeros of $r$.
Use symmetry to shorten the interval — then multiply by the number of identical pieces.
Integrate $\tfrac12r^2$, usually with the half-angle identities $\cos^2u=\tfrac12(1+\cos2u)$ and $\sin^2u=\tfrac12(1-\cos2u)$ (Section 5.1).
✏️ Example 1 — Area inside a cardioid
Find the area of the region enclosed by the cardioid $r=2+2\cos\theta$.
💡 Show solution▶
Limits $r\ge0$ for all $\theta$ and the cardioid is traced exactly once as $\theta$ goes from $0$ to $2\pi$.
Check With symmetry: the upper half is traced for $0\le\theta\le\pi$, and $2\cdot\tfrac12\displaystyle\int_0^{\pi}(2+2\cos\theta)^2d\theta=\Big[6\theta+8\sin\theta+\sin2\theta\Big]_0^{\pi}=6\pi$ ✓. In general the cardioid $r=a(1+\cos\theta)$ encloses area $\tfrac32\pi a^2$ — one and a half times the circle of radius $a$.
✏️ Example 2 — Area of one petal of a rose
Find the area of one petal of the rose $r=3\sin2\theta$, and the area enclosed by the whole rose.
💡 Show solution▶
Limits $r=0$ when $\sin2\theta=0$. Between the consecutive zeros $\theta=0$ and $\theta=\tfrac{\pi}{2}$ we have $r\ge0$, and the arm sweeps out the petal in quadrant I exactly once (Example 4 of Section 10.3).
Whole rose The four petals are congruent, so $A=4\cdot\tfrac{9\pi}{8}=\tfrac{9\pi}{2}\approx14.137$. Equivalently $\tfrac12\displaystyle\int_0^{2\pi}9\sin^22\theta\,d\theta=\tfrac{9\pi}{2}$ — here $0\le\theta\le2\pi$ is correct because $n=2$ is even and every petal is traced exactly once.
Check Interactive 1, “one petal”: with $n=120$ slices the inner sum $3.4754$ and the outer sum $3.5932$ trap $\tfrac{9\pi}{8}\approx3.5343$, and the gap halves each time $n$ doubles. ✓
✏️ Example 3 — The inner loop of a limaçon
Find the area enclosed by the inner loop of $r=1+2\cos\theta$.
💡 Show solution▶
Limits $r=0$ when $\cos\theta=-\tfrac12$: $\theta=\tfrac{2\pi}{3}$ and $\theta=\tfrac{4\pi}{3}$. Between them $r\le0$, and the arm (pointing backwards) sweeps out the inner loop exactly once. By the Remark after Theorem 10.8 the formula applies.
Check The answer is positive and small, as the picture of a little loop suggests; Interactive 1 (“inner loop”) gives sums that close in on $0.5435$. ✓
⚠️ Common mistakes with polar areas
Going round twice. The circle $r=2\cos\theta$ is traced once for $0\le\theta\le\pi$. Integrating over $[0,2\pi]$ gives $\tfrac12\int_0^{2\pi}4\cos^2\theta\,d\theta=2\pi$ — double the true area $\pi$. Always ask: over which interval is the region swept exactly once?
Forgetting the $\tfrac12$ or the square. The integrand is $\tfrac12r^2$, not $r$.
Using the limaçon's full period for one loop. $\tfrac12\int_0^{2\pi}(1+2\cos\theta)^2d\theta=3\pi$ counts the inner loop twice (once inside the outer loop, once on its own) — it is neither the outer area nor the inner one.
Area Between Two Polar Curves
If the region lies between an outer curve $r=f(\theta)$ and an inner curve $r=g(\theta)$, each thin slice is the difference of two sectors, with area approximately $\tfrac12f(\theta)^2\Delta\theta-\tfrac12g(\theta)^2\Delta\theta$.
📐 Theorem 10.9 — Area between two polar curves
Let $f$ and $g$ be continuous on $[\alpha,\beta]$ with $0\le g(\theta)\le f(\theta)$ there, and $0\lt\beta-\alpha\le2\pi$. The area of the region between $r=g(\theta)$ and $r=f(\theta)$ for $\alpha\le\theta\le\beta$ is
Note that this is $\tfrac12\int\left(f^2-g^2\right)d\theta$, not $\tfrac12\int(f-g)^2\,d\theta$: we subtract two sector areas.
🧭 Finding all the intersection points of two polar curves
Because one point has many polar names, solving $f(\theta)=g(\theta)$ can miss intersections. To find them all:
Solve $f(\theta)=g(\theta)$.
Check the pole separately. If $f(\theta_1)=0$ for some $\theta_1$ and $g(\theta_2)=0$ for some $\theta_2$, both curves pass through the pole — an intersection, even when $\theta_1\ne\theta_2$.
Where a curve uses negative $r$, also solve $f(\theta)=-g(\theta+\pi)$, since $(-r,\theta+\pi)$ names the same point as $(r,\theta)$.
Sketch both curves and compare: every crossing in the picture must appear in your list.
✏️ Example 4 — Inside one curve and outside another
Find the area of the region inside the circle $r=3\sin\theta$ and outside the cardioid $r=1+\sin\theta$.
💡 Show solution▶
Intersect $3\sin\theta=1+\sin\theta\iff\sin\theta=\tfrac12\iff\theta=\tfrac{\pi}{6}$ or $\tfrac{5\pi}{6}$ (both at $r=\tfrac32$). Both curves also meet at the pole, but that point is not on the boundary of our region. For $\tfrac{\pi}{6}\le\theta\le\tfrac{5\pi}{6}$ we have $3\sin\theta\ge1+\sin\theta\ge0$, so the circle is outer.
Set up $$A=\frac12\int_{\pi/6}^{5\pi/6}\left(9\sin^2\theta-(1+\sin\theta)^2\right)d\theta=\frac12\int_{\pi/6}^{5\pi/6}\left(8\sin^2\theta-2\sin\theta-1\right)d\theta.$$ With $8\sin^2\theta=4-4\cos2\theta$ the integrand is $3-4\cos2\theta-2\sin\theta$.
Check The circle has area $\pi\left(\tfrac32\right)^2=\tfrac{9\pi}{4}\approx7.07$, so a crescent of area $\pi$ is plausible. Interactive 2 (first option) accumulates the same total. ✓
✏️ Example 5 — A common region, and an intersection hiding at the pole
Find all points of intersection of the circles $r=2\cos\theta$ and $r=2\sin\theta$, and the area of the region inside both.
💡 Show solution▶
Intersect $2\cos\theta=2\sin\theta\iff\tan\theta=1\iff\theta=\tfrac{\pi}{4}$ (in $[0,\pi)$), giving the point $\left(\sqrt2,\tfrac{\pi}{4}\right)$, i.e. $(1,1)$. But the circles $(x-1)^2+y^2=1$ and $x^2+(y-1)^2=1$ clearly also meet at the origin. The pole is on $r=2\cos\theta$ at $\theta=\tfrac{\pi}{2}$ and on $r=2\sin\theta$ at $\theta=0$ — different angles, so the equation could not find it.
Describe As $\theta$ goes from $0$ to $\tfrac{\pi}{4}$, the boundary of the common region nearest the pole is $r=2\sin\theta$; from $\tfrac{\pi}{4}$ to $\tfrac{\pi}{2}$ it is $r=2\cos\theta$. The region is symmetric about the line $\theta=\tfrac{\pi}{4}$, so we double the first piece.
Check Geometry: the lens is two circular segments, each a quarter-disk ($\tfrac{\pi}{4}$) minus a right triangle ($\tfrac12$), giving $2\left(\tfrac{\pi}{4}-\tfrac12\right)$. ✓
Interactive 2 — Shading the Region Between Two Polar Curves
A ray sweeps from the first intersection angle to the last. Along the ray, the thin strip from the inner boundary to the outer boundary is added to the shaded region, contributing $\tfrac12\left(r_{\text{out}}^2-r_{\text{in}}^2\right)d\theta$. Dashed rays mark the intersection angles; a ringed pole marks an intersection that solving $f(\theta)=g(\theta)$ would miss.
$\theta$—
$r_{\text{in}}$—
$r_{\text{out}}$—
area shaded so far—
exact total—
Arc Length in Polar Coordinates
Treat $r=f(\theta)$ as the parametric curve $x=f(\theta)\cos\theta$, $y=f(\theta)\sin\theta$ and use the parametric arc length formula $L=\int\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}\,d\theta$ from Section 10.2. Squaring the derivatives found in Section 10.3,
$$\begin{aligned}\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2&=\left(f'\cos\theta-f\sin\theta\right)^2+\left(f'\sin\theta+f\cos\theta\right)^2\\&=f'^2\left(\cos^2\theta+\sin^2\theta\right)+f^2\left(\sin^2\theta+\cos^2\theta\right)=f^2+f'^2,\end{aligned}$$
because the cross terms $\mp2ff'\sin\theta\cos\theta$ cancel. This gives a remarkably tidy formula.
📐 Theorem 10.10 — Arc length of a polar curve
Let $f$ have a continuous derivative on $[\alpha,\beta]$, and suppose the curve $r=f(\theta)$ is traced exactly once as $\theta$ increases from $\alpha$ to $\beta$. Then its length is
Check: for the circle $r=a$ ($a\gt0$), $dr/d\theta=0$ and $L=\int_0^{2\pi}a\,d\theta=2\pi a$. ✓
✏️ Example 6 — The length of a cardioid, and of a polar parabola-like curve
(a) Show that the cardioid $r=a(1+\cos\theta)$, $a\gt0$, has length $8a$. (b) Find the length of the curve $r=\theta^2$ for $0\le\theta\le\sqrt5$.
💡 Show solution▶
(a) Simplify $r^2+r'^2=a^2(1+\cos\theta)^2+a^2\sin^2\theta=a^2\left(2+2\cos\theta\right)=4a^2\cos^2\tfrac{\theta}{2}$, using $1+\cos\theta=2\cos^2\tfrac{\theta}{2}$. So $\sqrt{r^2+r'^2}=2a\left|\cos\tfrac{\theta}{2}\right|$ — the absolute value matters!
(a) Integrate On $[0,2\pi]$, $\cos\tfrac{\theta}{2}\ge0$ only for $\theta\le\pi$. Using the symmetry about the polar axis, $$L=2\int_0^{\pi}2a\cos\tfrac{\theta}{2}\,d\theta=4a\Big[2\sin\tfrac{\theta}{2}\Big]_0^{\pi}=8a.$$ (Dropping the absolute value and integrating $2a\cos\tfrac{\theta}{2}$ over $[0,2\pi]$ gives $0$ — an impossible length.)
(b) $r'=2\theta$, so $r^2+r'^2=\theta^4+4\theta^2=\theta^2\left(\theta^2+4\right)$ and, for $\theta\ge0$, $\sqrt{r^2+r'^2}=\theta\sqrt{\theta^2+4}$. With $u=\theta^2+4$, $du=2\theta\,d\theta$: $$L=\int_0^{\sqrt5}\theta\sqrt{\theta^2+4}\,d\theta=\Big[\tfrac13\left(\theta^2+4\right)^{3/2}\Big]_0^{\sqrt5}=\tfrac13(27-8)=\frac{19}{3}\approx6.333.$$
Check Simpson's rule (Section 5.4) with $1000$ subintervals applied to $\int_0^{2\pi}\sqrt{r^2+r'^2}\,d\theta$ for $a=1$ gives $8.0000$, and for (b) gives $6.3333$. ✓
⚠️ Analysis — arc length pitfalls
1. Absolute values. $\sqrt{u^2}=|u|$, not $u$. Split the interval (or use symmetry) wherever the expression inside changes sign, as in Example 6(a).
2. Trace once only. The circle $r=2\sin\theta$ has $r^2+r'^2=4$; over $[0,\pi]$ the formula gives the correct $2\pi$, but over $[0,2\pi]$ it gives $4\pi$ because the circle is traced twice.
3. Most integrals have no elementary antiderivative. Even the limaçon $r=2+\cos\theta$ leads to $\int_0^{2\pi}\sqrt{5+4\cos\theta}\,d\theta$, which must be done numerically (Simpson's rule gives about $13.365$). Exact answers like $8a$ come from carefully chosen curves.
Exercises
I
Find the area of the region enclosed by each curve (or described region).
1 $r=4$
2 $r=6\sin\theta$
3 $r=3(1-\cos\theta)$
4 $r=2+\sin\theta$
5 $r=3+2\cos\theta$
6 both loops of the lemniscate $r^2=4\cos2\theta$
7 the region swept by the radius arm of the spiral $r=\theta$ as $\theta$ goes from $0$ to $\pi$
8 the region bounded by $r=\sec\theta$ and the rays $\theta=0$, $\theta=\tfrac{\pi}{4}$ (then check with elementary geometry)
💡 Show solutions▶
1 $\tfrac12\displaystyle\int_0^{2\pi}16\,d\theta=16\pi$ — the area of a disk of radius $4$. ✓
2 The circle is traced once on $[0,\pi]$: $\tfrac12\displaystyle\int_0^{\pi}36\sin^2\theta\,d\theta=9\displaystyle\int_0^{\pi}(1-\cos2\theta)\,d\theta=9\pi$ (radius $3$ ✓).
3 $r^2=9\left(1-2\cos\theta+\cos^2\theta\right)$; $\tfrac12\displaystyle\int_0^{2\pi}r^2d\theta=\tfrac92\left(2\pi+\pi\right)=\dfrac{27\pi}{2}\approx42.41$ (a cardioid with $a=3$: $\tfrac32\pi a^2$ ✓).
6 The right loop is swept for $-\tfrac{\pi}{4}\le\theta\le\tfrac{\pi}{4}$: $\tfrac12\displaystyle\int_{-\pi/4}^{\pi/4}4\cos2\theta\,d\theta=\Big[\sin2\theta\Big]_{-\pi/4}^{\pi/4}=2$. Two loops: $4$. (Use $r^2$ directly — no square root needed.)
8 $\tfrac12\displaystyle\int_0^{\pi/4}\sec^2\theta\,d\theta=\tfrac12\Big[\tan\theta\Big]_0^{\pi/4}=\tfrac12$. Geometry: $r=\sec\theta$ is the line $x=1$, so the region is the triangle with vertices $(0,0),(1,0),(1,1)$, area $\tfrac12$. ✓
II
Petals and loops.
1 one petal of $r=4\sin3\theta$
2 the whole rose $r=4\sin3\theta$ (careful with the interval!)
3 one petal of $r=2\cos4\theta$
4 the whole rose $r=2\cos4\theta$
5 the inner loop of $r=1+2\sin\theta$
6 the region between the inner and outer loops of $r=1+2\cos\theta$
2 $n=3$ is odd: three petals, traced once on $[0,\pi]$. Area $3\cdot\tfrac{4\pi}{3}=4\pi\approx12.566$. (Integrating over $[0,2\pi]$ would give $8\pi$ — every petal twice.)
3 Zeros at $\pm\tfrac{\pi}{8}$: $\tfrac12\displaystyle\int_{-\pi/8}^{\pi/8}4\cos^24\theta\,d\theta=\displaystyle\int_{-\pi/8}^{\pi/8}(1+\cos8\theta)\,d\theta=\dfrac{\pi}{4}\approx0.785$.
4 $n=4$ even: $8$ petals, $8\cdot\tfrac{\pi}{4}=2\pi$; equivalently $\tfrac12\displaystyle\int_0^{2\pi}4\cos^24\theta\,d\theta=2\pi$. (In general a rose has total area $\tfrac{\pi a^2}{2}$ for even $n$ and $\tfrac{\pi a^2}{4}$ for odd $n$.)
5 $r=0$ at $\theta=\tfrac{7\pi}{6},\tfrac{11\pi}{6}$; $r^2=3+4\sin\theta-2\cos2\theta$: $\tfrac12\Big[3\theta-4\cos\theta-\sin2\theta\Big]_{7\pi/6}^{11\pi/6}=\tfrac12\left(2\pi-4\sqrt3+\sqrt3\right)=\pi-\dfrac{3\sqrt3}{2}\approx0.5435$ — the same as Example 3, as rotation by $\tfrac{\pi}{2}$ predicts.
7 Loop in quadrant I, $0\le\theta\le\tfrac{\pi}{2}$: $\tfrac12\displaystyle\int_0^{\pi/2}9\sin2\theta\,d\theta=\tfrac92\Big[-\tfrac12\cos2\theta\Big]_0^{\pi/2}=\tfrac92$.
III
Regions between curves. Find all intersection points you need (remember the pole), then the area.
1 inside $r=3\cos\theta$ and outside $r=1+\cos\theta$
2 inside $r=1+\cos\theta$ and outside $r=1$
3 inside $r=1$ and outside $r=1+\cos\theta$
4 inside both $r=1$ and $r=1+\cos\theta$
5 inside both $r=\sqrt3\cos\theta$ and $r=\sin\theta$
6 inside the lemniscate $r^2=4\cos2\theta$ and outside the circle $r=\sqrt2$
7Challenge. Find all intersection points of $r=\cos\theta$ and $r=1-\cos\theta$, and the area inside both.
3 Now $\tfrac{\pi}{2}\le\theta\le\tfrac{3\pi}{2}$, where the circle is outer: $A=\tfrac12\displaystyle\int_{\pi/2}^{3\pi/2}\left(1-(1+\cos\theta)^2\right)d\theta=\tfrac12\displaystyle\int_{\pi/2}^{3\pi/2}\left(-2\cos\theta-\cos^2\theta\right)d\theta=\tfrac12\left(4-\tfrac{\pi}{2}\right)=2-\dfrac{\pi}{4}\approx1.215$.
4 Disk area minus part 3: $\pi-\left(2-\tfrac{\pi}{4}\right)=\dfrac{5\pi}{4}-2\approx1.927$. (Directly: $\tfrac12\int_{-\pi/2}^{\pi/2}1\,d\theta+\tfrac12\int_{\pi/2}^{3\pi/2}(1+\cos\theta)^2d\theta=\tfrac{\pi}{2}+\left(\tfrac{3\pi}{4}-2\right)$ ✓.)
5 $\sqrt3\cos\theta=\sin\theta\iff\tan\theta=\sqrt3\iff\theta=\tfrac{\pi}{3}$; both curves also pass through the pole ($\theta=0$ for $\sin\theta$, $\theta=\tfrac{\pi}{2}$ for $\sqrt3\cos\theta$). $A=\tfrac12\displaystyle\int_0^{\pi/3}\sin^2\theta\,d\theta+\tfrac12\displaystyle\int_{\pi/3}^{\pi/2}3\cos^2\theta\,d\theta=\left(\tfrac{\pi}{12}-\tfrac{\sqrt3}{16}\right)+\left(\tfrac{\pi}{8}-\tfrac{3\sqrt3}{16}\right)=\dfrac{5\pi}{24}-\dfrac{\sqrt3}{4}\approx0.2215$.
6 $4\cos2\theta=2\iff\cos2\theta=\tfrac12\iff\theta=\pm\tfrac{\pi}{6}$ (right loop; $\tfrac{5\pi}{6},\tfrac{7\pi}{6}$ for the left loop). Right loop: $\tfrac12\displaystyle\int_{-\pi/6}^{\pi/6}\left(4\cos2\theta-2\right)d\theta=\tfrac12\Big[2\sin2\theta-2\theta\Big]_{-\pi/6}^{\pi/6}=\sqrt3-\tfrac{\pi}{3}$. Both loops: $2\sqrt3-\dfrac{2\pi}{3}\approx1.370$.
7 $\cos\theta=1-\cos\theta\iff\cos\theta=\tfrac12$: the points $\left(\tfrac12,\pm\tfrac{\pi}{3}\right)$; also the pole ($r=\cos\theta$ at $\theta=\tfrac{\pi}{2}$, $r=1-\cos\theta$ at $\theta=0$). Near $\theta=0$ the cardioid is the inner boundary, for $\tfrac{\pi}{3}\le|\theta|\le\tfrac{\pi}{2}$ the circle is. By symmetry, $A=2\left[\tfrac12\displaystyle\int_0^{\pi/3}(1-\cos\theta)^2d\theta+\tfrac12\displaystyle\int_{\pi/3}^{\pi/2}\cos^2\theta\,d\theta\right]=\left(\tfrac{\pi}{2}-\tfrac{7\sqrt3}{8}\right)+\left(\tfrac{\pi}{12}-\tfrac{\sqrt3}{8}\right)=\dfrac{7\pi}{12}-\sqrt3\approx0.1005$.
IV
Arc length and applications.
1 Length of $r=5$, $0\le\theta\le2\pi$.
2 Length of $r=4\cos\theta$, $0\le\theta\le\pi$. Why not $0\le\theta\le2\pi$?
3 Length of $r=\theta^2$, $0\le\theta\le2\sqrt3$.
4 Length of $r=e^{2\theta}$, $0\le\theta\le\ln2$.
5 Length of the whole cardioid $r=2+2\cos\theta$.
6 Length of $r=\sin^3\!\left(\tfrac{\theta}{3}\right)$, $0\le\theta\le3\pi$.
7 Length of the first turn of the spiral $r=\theta$, $0\le\theta\le2\pi$ (use the trigonometric substitution $\theta=\tan u$, or Simpson's rule).
8Applied — radar coverage. Hills to the west limit a coastal radar's range, so it detects aircraft in the region $r\le20+10\cos\theta$ (km, $\theta$ measured from east). Find the area covered.
9Applied — a spiral shell. The edge of a snail shell follows the logarithmic spiral $r=e^{0.2\theta}$ cm for $0\le\theta\le4\pi$ (two turns). Find its length.
2 $r^2+r'^2=16\cos^2\theta+16\sin^2\theta=16$, so $L=\displaystyle\int_0^{\pi}4\,d\theta=4\pi$ — the circumference of a circle of diameter $4$ ✓. Over $[0,2\pi]$ the circle would be traced twice, giving $8\pi$.
3 As in Example 6(b): $L=\Big[\tfrac13\left(\theta^2+4\right)^{3/2}\Big]_0^{2\sqrt3}=\tfrac13(64-8)=\dfrac{56}{3}\approx18.67$.
6 $r'=\sin^2\!\tfrac{\theta}{3}\cos\tfrac{\theta}{3}$, so $r^2+r'^2=\sin^6\!\tfrac{\theta}{3}+\sin^4\!\tfrac{\theta}{3}\cos^2\!\tfrac{\theta}{3}=\sin^4\!\tfrac{\theta}{3}$ and $\sqrt{\ }=\sin^2\!\tfrac{\theta}{3}$ (never negative). $L=\displaystyle\int_0^{3\pi}\tfrac12\left(1-\cos\tfrac{2\theta}{3}\right)d\theta=\tfrac12\Big[\theta-\tfrac32\sin\tfrac{2\theta}{3}\Big]_0^{3\pi}=\dfrac{3\pi}{2}\approx4.712$.
8 $20+10\cos\theta\gt0$, traced once on $[0,2\pi]$: $A=\tfrac12\displaystyle\int_0^{2\pi}\left(400+400\cos\theta+100\cos^2\theta\right)d\theta=\tfrac12(800\pi+100\pi)=450\pi\approx1413.7$ km².
9 $r'=0.2e^{0.2\theta}$, $\sqrt{r^2+r'^2}=\sqrt{1.04}\,e^{0.2\theta}$: $L=\dfrac{\sqrt{1.04}}{0.2}\left(e^{0.8\pi}-1\right)\approx5.0990\times11.3453\approx57.85$ cm.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How long is the sheet wound on a roll of paper, steel or film — and can we find it without unrolling it?
Why it matters. Paper mills in Finland and Brazil, steel-coil plants in India and Germany, and cable manufacturers everywhere must know how much material is on a reel to invoice, schedule and avoid running out mid-job. Operators measure only the outer diameter; the length is then inferred from a model.
What to do.
Model the wound sheet (thickness $h$) on a core of radius $r_0$ as the Archimedean spiral $r=r_0+\dfrac{h}{2\pi}\theta$, $0\le\theta\le2\pi N$, where $N=\dfrac{R-r_0}{h}$ turns fill out to the outer radius $R$.
Write the exact length with Theorem 10.10 and evaluate it numerically (Simpson's rule or a spreadsheet) for a paper roll with $r_0=0.05$ m, $R=0.5$ m, $h=0.2$ mm.
Compare with the quick “area” estimate $L\approx\dfrac{\pi\left(R^2-r_0^2\right)}{h}$: the cross-section (an annulus) divided by the sheet thickness.
Explain why the two answers are so close: compare the sizes of $r^2$ and $\left(\dfrac{dr}{d\theta}\right)^2$ in the integrand.
Tools: Calculator, spreadsheet or a short program; optionally a roll of kitchen paper, a ruler and the sheet count on the packet.
Hand in: A half-page note with both lengths, the percentage difference, and one physical effect the model ignores (e.g. compression of the inner layers).
Going further. Measure a real household roll (core and outer diameter, thickness from the packet's sheet count and length) and test the model against the length printed on the packet.
Instructor note: $N=2250$ turns; both methods give about $3887.7$ m, differing by less than a millimetre, because $\left(\tfrac{h}{2\pi}\right)^2\approx10^{-9}$ m² is negligible beside $r^2\ge0.0025$ m², so $\sqrt{r^2+r'^2}\approx r$ and $\int r\,d\theta$ equals the annulus area divided by $h$ exactly.
🧪 Self-check quiz
0/5 correct
Q1. The area of the region bounded by $r=f(\theta)\ge0$, $\alpha\le\theta\le\beta$, is
Q2. The area enclosed by the circle $r=2\cos\theta$ is
Q3. The area of one petal of $r=\cos2\theta$ is
Q4. The length of the cardioid $r=3(1+\cos\theta)$ is
Q5. Solving $\cos\theta=\sin\theta$ finds only one intersection of $r=\cos\theta$ and $r=\sin\theta$. The other one is