Calculus 2

Area in polar coordinates

A polar curve is drawn by a distance that changes as the direction turns. How much area does it sweep out, and how much lies between two such curves?

You need

Systems with a curveAlgebra 1

Where you are 5 in Parametric, polar and vector valued functions

Calculus8

Parametric, polar and vector valued functions5

Area in polar coordinates

Derivatives of parametric curves Slope of a polar curve Arc length and speed on parametric and polar curves Vector valued functions and motion in the plane

When you reach for it

A region whose edge is a polar curve, such as a cardioid, a petal of a rose or a loop of a limaçon (Limacons and cardioids, Rose curves, Polar circles and symmetry tests), is awkward to cut into vertical strips: its rectangular equation is often of high degree and the top and bottom of a strip change formula partway across. The natural pieces for a polar region are thin wedges with their point at the pole. The area of a wedge of a circle is \displaystyle \frac{1}{2}r^2\theta (Sectors, segments and their areas), and adding wedges under a curve leads to one integral in \theta. Reach for it whenever the edge of the region is given as r = f(\theta), or when two polar curves bound a region between them, as in Area between curves for curves y = f(x).

The procedure

Day 1: the area swept by one curve

The formula from thin sectors. Let f be continuous with f(\theta) \ge 0 on [\alpha, \beta], where 0 < \beta - \alpha \le 2\pi. The region R is every point (r, \theta) with \alpha \le \theta \le \beta and 0 \le r \le f(\theta): the region swept by the segment from the pole to the curve as the direction turns from \alpha to \beta.

Cut [\alpha, \beta] into n pieces of width \displaystyle \Delta\theta = \frac{\beta - \alpha}{n}. Over one piece, the slice of R lies between two circular sectors with the same angle \Delta\theta: one whose radius is the smallest value m_i of f on the piece, and one whose radius is the largest value M_i. So the slice has area between \displaystyle \frac{1}{2}m_i^2\,\Delta\theta and \displaystyle \frac{1}{2}M_i^2\,\Delta\theta. Adding over all the pieces,

\begin{aligned} \sum_{i=1}^{n} \tfrac{1}{2}m_i^2\,\Delta\theta &\le \text{Area}(R) \\ &\le \sum_{i=1}^{n} \tfrac{1}{2}M_i^2\,\Delta\theta. \end{aligned}

Both outer sums are Riemann sums of the continuous function \displaystyle \frac{1}{2}f(\theta)^2, the lower and upper ones, so as n \to \infty both tend to the same The definite integral. The area is squeezed between them:

A = \frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta.

Figure 1Six thin sectors under r = 5 + 3\sin\theta for \displaystyle 0 \le \theta \le \frac{\pi}{2}. As the angles shrink, their total area tends to \displaystyle \frac{1}{2}\int r^2\,d\theta.

The factor \displaystyle \frac{1}{2} and the square come straight from the sector formula. For a constant r = R the integral gives \displaystyle \frac{1}{2}R^2(\beta - \alpha), the sector itself.

Evaluating it. The integrand is a square, so expanding it usually leaves \cos^2 or \sin^2 of a multiple of \theta. Power reduction (Trigonometric integrals) turns those into first powers:

\cos^2 u = \frac{1 + \cos 2u}{2}, \qquad \sin^2 u = \frac{1 - \cos 2u}{2}.

The integral is then finished with an antiderivative (The second fundamental theorem of calculus). Over a full turn, \int_0^{2\pi} \cos k\theta\,d\theta = \int_0^{2\pi} \sin k\theta\,d\theta = 0 for every positive integer k, and \int_0^{2\pi} \cos^2 k\theta\,d\theta and \int_0^{2\pi} \sin^2 k\theta\,d\theta both equal \pi.

Choosing the angles. The integral adds up whatever the segment from the pole sweeps, once for each time it sweeps it. So the interval of \theta must draw the region’s edge exactly once, and nothing else:

For a whole closed curve that never passes through the pole, one full turn, [0, 2\pi].
For one petal or one loop, the interval between two neighboring zeros of r, since the curve leaves the pole and returns to it there.
For a curve drawn once in half a turn, such as an odd rose or a circle through the pole, [0, \pi]. A full turn would sweep every point twice and double the answer.

A piece drawn with r < 0 lies on the far side of the pole, in the direction \theta + \pi, but at distance |r|, and r^2 = |r|^2. So the formula gives that piece’s area correctly; the sign of r only says where the piece is.

Day 2: the area between two curves

The formula. If 0 \le r_{\text{in}}(\theta) \le r_{\text{out}}(\theta) for \alpha \le \theta \le \beta, the region between the two curves is what the outer curve sweeps with what the inner curve sweeps taken away, so

A = \frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\text{out}}^2 - r_{\text{in}}^2\right) d\theta.

Each radius is squared before subtracting: the region is a difference of two sector-shaped areas, not a sector of radius r_{\text{out}} - r_{\text{in}}.

Finding the angles. The ends of the interval are usually where the curves cross. Setting f(\theta) = g(\theta) is a system with a curve in polar form, and it finds the crossings where both curves reach the point at the same \theta. It can miss two kinds of crossing:

The pole. The pole is (0, \theta) for every \theta. Two curves can pass through it at different angles, so it never solves f(\theta) = g(\theta). Test it separately: the pole is on r = f(\theta) exactly when f(\theta) = 0 for some \theta.
Other names for a point. A point is also (-r, \theta + \pi). Where one curve has negative r, also solve f(\theta) = -g(\theta + \pi), as in the rose crossing a circle in Rose curves.

A sketch decides which curve is outside on each piece. Where the outer and inner curves trade places, the integral splits there, and a region bounded by the pole on one side uses the one-curve formula on that piece.

Worked examples

Example 1

Day 1: Altantsetseg’s sprinkler

A lawn sprinkler sits at the pole. Altantsetseg measures the edge of the wet ground and finds it is r = 5 + 3\sin\theta, in feet. Find the area of the wet ground.

Solution

Since \sin\theta \ge -1, the distance r \ge 2 > 0, so the curve never reaches the pole and one turn, 0 \le \theta \le 2\pi, draws it once (Limacons and cardioids: a limaçon without an inner loop). Expand the square:

r^2 = 25 + 30\sin\theta + 9\sin^2\theta.

Over a full turn, \int_0^{2\pi} \sin\theta\,d\theta = 0 and \int_0^{2\pi} \sin^2\theta\,d\theta = \pi. So

A = \frac{1}{2}\left(25 \cdot 2\pi + 0 + 9\pi\right) = \frac{59\pi}{2} \approx 92.7.

The wet ground covers \displaystyle \frac{59\pi}{2} \approx 92.7 square feet. As a check on size, a circle of radius 5, the average distance, has area 25\pi \approx 78.5; the extra comes from squaring, since the far side gains more than the near side loses.

Example 2

Day 1: one petal of an eight-petal rose

Find the area of one petal of r = 8\cos 4\theta, and of the whole rose.

Solution

The petal on the polar axis runs between the zeros of \cos 4\theta nearest 0: \displaystyle 4\theta = \pm\frac{\pi}{2}, so \displaystyle -\frac{\pi}{8} \le \theta \le \frac{\pi}{8}. On this interval r \ge 0, and the segment from the pole sweeps the petal once. Using \displaystyle \cos^2 4\theta = \frac{1 + \cos 8\theta}{2},

\begin{aligned} A &= \frac{1}{2}\int_{-\pi/8}^{\pi/8} 64\cos^2 4\theta\,d\theta \\ &= 16\int_{-\pi/8}^{\pi/8} \left(1 + \cos 8\theta\right) d\theta \\ &= 16\left[\theta + \frac{\sin 8\theta}{8}\right]_{-\pi/8}^{\pi/8} = 16 \cdot \frac{\pi}{4} = 4\pi. \end{aligned}

The term \sin 8\theta is 0 at both ends. Here n = 4 is even, so the rose has 8 petals of the same size, and the whole rose has area 8 \cdot 4\pi = 32\pi. The same number comes from one full turn, \displaystyle \frac{1}{2}\int_0^{2\pi} 64\cos^2 4\theta\,d\theta = 32\pi, because an even rose is drawn once on [0, 2\pi]. It is half the area 64\pi of the circle r = 8 that the tips touch.

Example 3

Day 1: a five-petal rose and the interval that counts it once

Find the total area enclosed by r = 4\sin 5\theta, and explain why \displaystyle \frac{1}{2}\int_0^{2\pi} r^2\,d\theta gives the wrong number.

Solution

Here n = 5 is odd, so the rose has 5 petals and is drawn once for 0 \le \theta \le \pi (Rose curves). The five arches of \sin 5\theta on [0, \pi] draw the five petals, some with r < 0, and r^2 counts each by its true size. Using \displaystyle \sin^2 5\theta = \frac{1 - \cos 10\theta}{2},

\begin{aligned} A &= \frac{1}{2}\int_0^{\pi} 16\sin^2 5\theta\,d\theta \\ &= 4\left[\theta - \frac{\sin 10\theta}{10}\right]_0^{\pi} = 4\pi. \end{aligned}

Over [0, 2\pi] the integral is 8\pi. The reason is that \sin 5(\theta + \pi) = -\sin 5\theta, so the point at \theta + \pi is (-r, \theta + \pi), which is the same point as (r, \theta). The second half turn draws all five petals a second time, and the integral adds every petal twice.

Example 4

Day 1: the inner loop of a limaçon

Find the area inside the inner loop of r = 4 + 8\sin\theta.

Solution

The loop starts and ends where r = 0: \displaystyle \sin\theta = -\frac{1}{2}, so \displaystyle \theta = \frac{7\pi}{6} and \displaystyle \theta = \frac{11\pi}{6}. Between them \displaystyle \sin\theta < -\frac{1}{2} and r < 0, and the point traces the small loop above the pole, starting and ending at the pole.

Figure 2Example 4. The inner loop of r = 4 + 8\sin\theta, drawn while \displaystyle \frac{7\pi}{6} \le \theta \le \frac{11\pi}{6} and r \le 0.

Expand: r^2 = 16 + 64\sin\theta + 64\sin^2\theta, which by power reduction is 48 + 64\sin\theta - 32\cos 2\theta. An antiderivative is 48\theta - 64\cos\theta - 16\sin 2\theta. At the two ends, \displaystyle \cos\frac{11\pi}{6} - \cos\frac{7\pi}{6} = \sqrt{3} and \displaystyle \sin\frac{11\pi}{3} - \sin\frac{7\pi}{3} = -\sqrt{3}. So

\begin{aligned} A &= \frac{1}{2}\left[48 \cdot \frac{2\pi}{3} - 64\sqrt{3} + 16\sqrt{3}\right] \\ &= \frac{1}{2}\left(32\pi - 48\sqrt{3}\right) \\ &= 16\pi - 24\sqrt{3} \approx 8.70. \end{aligned}

The loop reaches 4 units from the pole (at \displaystyle \theta = \frac{3\pi}{2}, r = -4), so it fits inside a circle of radius 4 and area 16\pi \approx 50.3; an area near 9 is the right size for a loop that thin.

Example 5

Day 2: Ausra’s microphone beyond six feet

Ausra’s microphone sits at the pole facing along the polar axis, and it picks up a voice anywhere inside the cardioid r = 4 + 4\cos\theta, in feet. A stage edge runs along the circle r = 6. Find the area of the pickup region that lies beyond the circle.

Solution

The crossings: 4 + 4\cos\theta = 6 gives \displaystyle \cos\theta = \frac{1}{2}, so \displaystyle \theta = \pm\frac{\pi}{3}. The cardioid reaches the pole (at \theta = \pi) but the circle does not, and both have r \ge 0 everywhere, so these two are the only crossings. Between them, 4 + 4\cos\theta \ge 6: the cardioid is the outer curve.

Figure 3Example 5. The pickup region beyond 6 feet lies inside the cardioid and outside the circle, for \displaystyle -\frac{\pi}{3} \le \theta \le \frac{\pi}{3}.

A = \frac{1}{2}\int_{-\pi/3}^{\pi/3} \left[(4 + 4\cos\theta)^2 - 36\right] d\theta.

Expand: (4 + 4\cos\theta)^2 - 36 is 16 + 32\cos\theta + 16\cos^2\theta - 36, which is -12 + 32\cos\theta + 8\cos 2\theta. An antiderivative is -12\theta + 32\sin\theta + 4\sin 2\theta, an odd function, so its change over \displaystyle \left[-\frac{\pi}{3}, \frac{\pi}{3}\right] is twice its value at \displaystyle \frac{\pi}{3}:

\begin{aligned} A &= \frac{1}{2} \cdot 2\left(-4\pi + 16\sqrt{3} + 2\sqrt{3}\right) \\ &= 18\sqrt{3} - 4\pi \approx 18.6. \end{aligned}

About 18.6 square feet of the pickup region lies beyond the stage edge.

Example 6

Day 2: the region inside both, and a crossing at the pole

Find every point where the cardioids r = 2 + 2\cos\theta and r = 2 - 2\sin\theta meet, and the area of the region inside both.

Solution

Setting the two equal, 2 + 2\cos\theta = 2 - 2\sin\theta, so \cos\theta = -\sin\theta, \tan\theta = -1, and \displaystyle \theta = \frac{3\pi}{4} or \displaystyle \theta = \frac{7\pi}{4}: the points \displaystyle \left(2 - \sqrt{2}, \frac{3\pi}{4}\right) and \displaystyle \left(2 + \sqrt{2}, \frac{7\pi}{4}\right). The pole needs its own test. The first cardioid passes through it at \theta = \pi, where \cos\theta = -1, and the second at \displaystyle \theta = \frac{\pi}{2}, where \sin\theta = 1. Both curves contain the pole, at different angles, so it is a third crossing that the equation could not find. Both have r \ge 0 for every \theta, so no crossing hides under another name.

Figure 4Example 6. The region inside both cardioids. The curves meet at three points, and the pole is not a solution of 2 + 2\cos\theta = 2 - 2\sin\theta.

For a direction \theta, a point is inside both curves when its distance is below both radii, so the region is bounded by the smaller one. Since \displaystyle \cos\theta + \sin\theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right), which is \ge 0 for \displaystyle -\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}, there 2 - 2\sin\theta \le 2 + 2\cos\theta and the second cardioid is the edge; for \displaystyle \frac{3\pi}{4} \le \theta \le \frac{7\pi}{4} the first one is. Reflection in the line y = -x sends the direction \theta to \displaystyle \frac{3\pi}{2} - \theta, and \displaystyle 2 + 2\cos\left(\frac{3\pi}{2} - \theta\right) = 2 - 2\sin\theta, so the reflection swaps the two cardioids and the two pieces have equal area:

A = 2 \cdot \frac{1}{2}\int_{-\pi/4}^{3\pi/4} (2 - 2\sin\theta)^2\,d\theta.

Expand: (2 - 2\sin\theta)^2 = 6 - 8\sin\theta - 2\cos 2\theta, with antiderivative 6\theta + 8\cos\theta - \sin 2\theta. That is \displaystyle \frac{9\pi}{2} - 4\sqrt{2} + 1 at \displaystyle \frac{3\pi}{4} and \displaystyle -\frac{3\pi}{2} + 4\sqrt{2} + 1 at \displaystyle -\frac{\pi}{4}, so

A = 6\pi - 8\sqrt{2} \approx 7.54.

Each cardioid alone encloses 6\pi \approx 18.8, so a shared part of about 7.5 is the right size.

Example 7

Day 2: between the two loops of a limaçon

Find the area of the region inside the outer loop of r = 4 + 8\sin\theta and outside its inner loop.

Solution

One full turn draws both loops once: the outer loop while r \ge 0, for \displaystyle -\frac{\pi}{6} \le \theta \le \frac{7\pi}{6}, and the inner loop while r \le 0 (Example 4). So \displaystyle \frac{1}{2}\int_0^{2\pi} r^2\,d\theta adds the two regions:

\begin{aligned} \frac{1}{2}\int_0^{2\pi} (4 + 8\sin\theta)^2\,d\theta &= \frac{1}{2}\left(16 \cdot 2\pi + 0 + 64\pi\right) \\ &= 48\pi. \end{aligned}

With the inner loop’s area \displaystyle 16\pi - 24\sqrt{3} from Example 4, the outer loop encloses \displaystyle 48\pi - \left(16\pi - 24\sqrt{3}\right) = 32\pi + 24\sqrt{3}. The inner loop lies inside the outer one, so the region between them is

\begin{aligned} &\left(32\pi + 24\sqrt{3}\right) - \left(16\pi - 24\sqrt{3}\right) \\ &= 16\pi + 48\sqrt{3} \approx 133.4. \end{aligned}

A check by direction: for \displaystyle \frac{\pi}{6} \le \theta \le \frac{5\pi}{6} the inner loop lies at distance 8\sin\theta - 4 (its points at \theta + \pi with r < 0, turned around), so there the region runs from 8\sin\theta - 4 out to 4 + 8\sin\theta, and (4 + 8\sin\theta)^2 - (8\sin\theta - 4)^2 = 128\sin\theta. For \displaystyle -\frac{\pi}{6} \le \theta \le \frac{\pi}{6} and \displaystyle \frac{5\pi}{6} \le \theta \le \frac{7\pi}{6} only the outer loop is present, in two pieces that are mirror images across the line \displaystyle \theta = \frac{\pi}{2}. So

\begin{aligned} A &= \frac{1}{2}\int_{\pi/6}^{5\pi/6} 128\sin\theta\,d\theta \\ &\quad + 2 \cdot \frac{1}{2}\int_{-\pi/6}^{\pi/6} (4 + 8\sin\theta)^2\,d\theta \\ &= 64\sqrt{3} + \left(16\pi - 16\sqrt{3}\right), \end{aligned}

the same \displaystyle 16\pi + 48\sqrt{3}.

Boundaries

The interval must sweep the region once. The formula adds area every time the segment from the pole passes over it. If \beta - \alpha > 2\pi, or the curve retraces itself on the interval (an odd rose or a circle through the pole over a full turn), the integral counts some area twice.

Each ray from the pole must leave the region once. The one-curve formula covers regions of the form 0 \le r \le f(\theta). A region not touching the pole needs the two-curve formula, and a region whose outer edge changes formula needs the integral split where it changes.

The inner curve must stay inside. \displaystyle \frac{1}{2}\int \left(r_{\text{out}}^2 - r_{\text{in}}^2\right) d\theta needs 0 \le r_{\text{in}} \le r_{\text{out}} on the whole interval. If the curves cross inside the interval the integrand changes sign and parts cancel.

Solving f(\theta) = g(\theta) is not the whole list of crossings. The pole and points named with negative r must be checked apart from the equation.

Continuity. The derivation needs f continuous on [\alpha, \beta], so that the upper and lower sector sums close in on one number.

Where this goes next

The same thin wedges measure a curve’s length: Arc length and speed on parametric and polar curves finds the arc length of a polar curve from \displaystyle \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}. The direction of the curve at each point, and at the pole in particular, comes from Slope of a polar curve.

Exercises

Day 1: area inside one curve

Find the area of the region inside r = 7 with \displaystyle 0 \le \theta \le \frac{\pi}{3}, by the integral.
Find the area inside r = 6 - 2\sin\theta.
Find the area inside the cardioid r = 7 - 7\cos\theta.
Find the area of the region swept by the spiral r = e^{\theta/2} for 0 \le \theta \le \pi.
Find the area of the part of the circle r = 6\sin\theta with \displaystyle \frac{\pi}{4} \le \theta \le \frac{\pi}{2}, and check it with plane geometry.

Day 1: petals and loops

Find the area of one petal of r = 5\sin 6\theta, and of the whole rose.
Find the area of one petal of r = 10\cos 3\theta, and of the whole rose.
Find the total area enclosed by r = 7\cos 2\theta.
Find the area inside the inner loop of \displaystyle r = 2 + 2\sqrt{2}\sin\theta.

Day 1: reasoning

The circle r = 2\cos\theta has radius 1, yet \displaystyle \frac{1}{2}\int_0^{2\pi} 4\cos^2\theta\,d\theta = 2\pi. Explain the difference and give an interval that gives the true area.
Show that for a constant r = R, the area formula gives the area of a sector, and use it to find the area of a circle of radius R.
Show that \displaystyle \frac{1}{2}\int_0^{2\pi} (a + b\cos\theta)^2\,d\theta equals \displaystyle \pi\left(a^2 + \frac{b^2}{2}\right), and explain what area it measures when 0 < b \le a and when 0 < a < b.

Day 2: between two curves

Find the area inside the circle \displaystyle r = 4\sqrt{3}\sin\theta and outside the circle r = 6.
Find the area of the region inside both \displaystyle r = 4\sqrt{3}\sin\theta and r = 6.
Erdene’s lawn sprinkler, at the pole, wets the ground inside r = 8 + 8\sin\theta, in feet. Find the area of wet ground more than 8 feet from the sprinkler.
Find the area inside the circle r = 3 and outside the cardioid r = 3 - 3\cos\theta.
Find every point where r = 6 + 6\sin\theta meets r = 12\sin\theta, and the area inside the cardioid and outside the circle.
Find the area between the two loops of \displaystyle r = 2 + 2\sqrt{2}\sin\theta.

Day 2: where the curves meet

Find every point where \displaystyle r = 2\sqrt{3}\sin\theta meets r = 2\cos\theta, and the area of the region inside both.
Find the area inside the rose r = 4\sin 3\theta and outside the circle r = 2.

Day 2: reasoning

Explain why the area between two polar curves is \displaystyle \frac{1}{2}\int \left(r_{\text{out}}^2 - r_{\text{in}}^2\right) d\theta and not \displaystyle \frac{1}{2}\int \left(r_{\text{out}} - r_{\text{in}}\right)^2 d\theta, using the ring between r = 1 and r = 3.
Explain why solving f(\theta) = g(\theta) can miss the pole as a crossing, and how to decide whether two polar curves both pass through the pole.

Answers

\displaystyle \frac{49\pi}{6}. \displaystyle \frac{1}{2}\int_0^{\pi/3} 49\,d\theta = \frac{49}{2} \cdot \frac{\pi}{3}.
38\pi. r^2 = 36 - 24\sin\theta + 4\sin^2\theta, and r \ge 4, so one turn: \displaystyle \frac{1}{2}\left(72\pi + 0 + 4\pi\right).
\displaystyle \frac{147\pi}{2}. r^2 = 49\left(1 - 2\cos\theta + \cos^2\theta\right); over [0, 2\pi] the integral is 49(2\pi + \pi) = 147\pi, then halve.
\displaystyle \frac{e^{\pi} - 1}{2} \approx 11.07. r^2 = e^{\theta}, so \displaystyle \frac{1}{2}\int_0^{\pi} e^{\theta}\,d\theta.
\displaystyle \frac{9\pi}{4} + \frac{9}{2}. The integral \displaystyle \frac{1}{2}\int_{\pi/4}^{\pi/2} 36\sin^2\theta\,d\theta is \displaystyle 9\left[\theta - \frac{\sin 2\theta}{2}\right]_{\pi/4}^{\pi/2}, which is \displaystyle 9\left(\frac{\pi}{4} + \frac{1}{2}\right). Geometry: the circle has center (0, 3) and radius 3; the region is a quarter of the disk (\displaystyle \frac{9\pi}{4}) plus the right triangle with vertices the pole, (0, 3) and (3, 3) (\displaystyle \frac{9}{2}).
One petal \displaystyle \frac{25\pi}{24}, whole rose \displaystyle \frac{25\pi}{2}. A petal runs over \displaystyle 0 \le \theta \le \frac{\pi}{6}, and \displaystyle \frac{25}{2}\int_0^{\pi/6} \sin^2 6\theta\,d\theta is \displaystyle \frac{25}{2} \cdot \frac{\pi}{12}. The rose has 12 petals.
One petal \displaystyle \frac{25\pi}{3}, whole rose 25\pi. The petal runs over \displaystyle -\frac{\pi}{6} \le \theta \le \frac{\pi}{6}: \displaystyle 50\int_{-\pi/6}^{\pi/6} \cos^2 3\theta\,d\theta = 50 \cdot \frac{\pi}{6}. There are 3 petals.
\displaystyle \frac{49\pi}{2}. The rose has 4 petals and is drawn once on [0, 2\pi]: \displaystyle \frac{49}{2}\int_0^{2\pi} \cos^2 2\theta\,d\theta = \frac{49}{2}\pi.
2\pi - 6 \approx 0.28. r = 0 at \displaystyle \sin\theta = -\frac{\sqrt{2}}{2}: \displaystyle \theta = \frac{5\pi}{4}, \frac{7\pi}{4}. \displaystyle r^2 = 8 + 8\sqrt{2}\sin\theta - 4\cos 2\theta, antiderivative \displaystyle 8\theta - 8\sqrt{2}\cos\theta - 2\sin 2\theta, which changes by 4\pi - 16 + 4; halve it.
The circle is drawn once on [0, \pi] and again on [\pi, 2\pi] (there r < 0 and the points repeat), so the full turn counts it twice. On [0, \pi] the integral \displaystyle \frac{1}{2}\int_0^{\pi} 4\cos^2\theta\,d\theta is \pi, the area of a circle of radius 1.
\displaystyle \frac{1}{2}\int_{\alpha}^{\beta} R^2\,d\theta = \frac{1}{2}R^2(\beta - \alpha), the sector formula with angle \beta - \alpha. With [0, 2\pi] this is \pi R^2.
Expand: a^2 + 2ab\cos\theta + b^2\cos^2\theta; the middle term integrates to 0 and the last to b^2\pi, so the integral is \displaystyle \frac{1}{2}\left(2\pi a^2 + \pi b^2\right). When b \le a, r \ge 0 and the curve is drawn once with no inner loop, so this is the area it encloses. When b > a, the full turn draws the outer loop (where r > 0) and the inner loop (where r < 0), so this is the outer loop’s area plus the inner loop’s area; the inner loop sits inside the outer one, so its area is counted twice compared with the region the curve encloses.
\displaystyle 6\sqrt{3} - 2\pi \approx 4.11. \displaystyle 4\sqrt{3}\sin\theta = 6 at \displaystyle \theta = \frac{\pi}{3}, \frac{2\pi}{3}; the area is \displaystyle \frac{1}{2}\int_{\pi/3}^{2\pi/3} \left(48\sin^2\theta - 36\right) d\theta, which is \displaystyle \frac{1}{2}\left(8\pi + 12\sqrt{3} - 12\pi\right).
\displaystyle 14\pi - 6\sqrt{3} \approx 33.59. The circle \displaystyle r = 4\sqrt{3}\sin\theta has radius \displaystyle 2\sqrt{3} and area 12\pi; take away the part outside r = 6 from exercise 13.
128 + 16\pi \approx 178.3 square feet. 8 + 8\sin\theta = 8 at \theta = 0, \pi, and the cardioid is outside for 0 < \theta < \pi. The area is \displaystyle \frac{1}{2}\int_0^{\pi} 128\sin\theta\,d\theta plus \displaystyle \frac{1}{2}\int_0^{\pi} 64\sin^2\theta\,d\theta, which is \displaystyle \frac{1}{2}\left(256 + 32\pi\right).
\displaystyle 18 - \frac{9\pi}{4} \approx 10.93. They meet at \displaystyle \theta = \pm\frac{\pi}{2}, and the area is \displaystyle \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left(18\cos\theta - 9\cos^2\theta\right) d\theta, which is \displaystyle \frac{1}{2}\left(36 - \frac{9\pi}{2}\right).
They meet at \displaystyle \left(12, \frac{\pi}{2}\right) and at the pole (the circle at \theta = 0, the cardioid at \displaystyle \theta = \frac{3\pi}{2}). Area 18\pi. The circle is drawn once for 0 \le \theta \le \pi, and there 6 + 6\sin\theta \ge 12\sin\theta because \sin\theta \le 1, so the circle lies inside the cardioid: \displaystyle \frac{3\pi \cdot 36}{2} - 36\pi = 54\pi - 36\pi.
4\pi + 12 \approx 24.57. A full turn gives \displaystyle \frac{1}{2}\int_0^{2\pi} r^2\,d\theta, which is \displaystyle \frac{1}{2}\left(8\pi + 8\pi\right) = 8\pi, the two loops together. Outer loop 8\pi - (2\pi - 6) = 6\pi + 6, then subtract the inner loop 2\pi - 6 (exercise 9).
\displaystyle \left(\sqrt{3}, \frac{\pi}{6}\right) and the pole; area \displaystyle \frac{5\pi}{6} - \sqrt{3} \approx 0.89. \displaystyle \tan\theta = \frac{1}{\sqrt{3}} gives \displaystyle \frac{\pi}{6}; the pole is on both (\theta = 0 and \displaystyle \theta = \frac{\pi}{2}). The sine circle is inside for \displaystyle 0 \le \theta \le \frac{\pi}{6} and the cosine circle after, so the area is \displaystyle \frac{1}{2}\int_0^{\pi/6} 12\sin^2\theta\,d\theta plus \displaystyle \frac{1}{2}\int_{\pi/6}^{\pi/2} 4\cos^2\theta\,d\theta, which is \displaystyle \left(\frac{\pi}{2} - \frac{3\sqrt{3}}{4}\right) + \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right).
\displaystyle \frac{4\pi}{3} + 2\sqrt{3} \approx 7.65. On the petal \displaystyle 0 \le \theta \le \frac{\pi}{3}, 4\sin 3\theta \ge 2 for \displaystyle \frac{\pi}{18} \le \theta \le \frac{5\pi}{18}. On one petal the area is \displaystyle \frac{1}{2}\int_{\pi/18}^{5\pi/18} \left(16\sin^2 3\theta - 4\right) d\theta, which is \displaystyle \frac{4\pi}{9} + \frac{2\sqrt{3}}{3}. The three petals are congruent.
The ring has area 9\pi - \pi = 8\pi, and \displaystyle \frac{1}{2}\int_0^{2\pi} (9 - 1)\,d\theta = 8\pi agrees. The other form gives \displaystyle \frac{1}{2}\int_0^{2\pi} 4\,d\theta = 4\pi, the area of a disk whose radius is the ring’s width 2: it moves the ring in to the pole, where wedges are narrowest. The region is the outer swept area minus the inner one, and each is \displaystyle \frac{1}{2}\int r^2\,d\theta.
The pole is (0, \theta) for every \theta, so a curve can reach it at an angle where the other curve is elsewhere; f(\theta) = g(\theta) asks for one \theta that works for both. Decide it by solving f(\theta) = 0 and g(\theta) = 0 separately: if each has a solution, both curves pass through the pole.