The limits are where the curve finishes BC only

Sweeping the angle by $d\theta$ sweeps out a thin circular sector of radius $r$, whose area is $\frac{1}{2}r^{2}\,d\theta$. Adding them gives

$$A = \frac{1}{2}\int_{\alpha}^{\beta} \big(f(\theta)\big)^{2}\,d\theta.$$

Both constants go missing independently of each other and independently of the limits, so a student can get the interval exactly right and still be wrong by a factor of two:

1. The $\frac{1}{2}$ dropped: every answer is doubled.
2. The square dropped: $\frac{1}{2}\int r\,d\theta$ accumulates lengths rather than areas, and does not have the units of an area at all.
3. Squared after integrating: $\frac{1}{2}\left(\int r\,d\theta\right)^{2}$, the same misplacement as in 8.9's disc integral. Each sector uses its own radius.

One sanity check settles all three at once. Compare the answer with a circle you can compute by hand. The curve $r = 2\cos\theta$ is a circle of radius $1$, so its area must be $\pi$; any setup returning $2\pi$ or $\frac{4}{\pi}$ has a constant misplaced, and no algebra is needed to see it.

A petal of a rose starts and ends at the origin, so its angular limits are consecutive solutions of $r = 0$. That is the procedure, and it is short:

1. Set $f(\theta) = 0$ and solve. These are the angles at which the curve passes through the pole.
2. Take two consecutive roots. Between them the curve is out at a nonzero radius, tracing exactly one loop.
3. Integrate between those two. One petal, once.

For $r = 2\sin 3\theta$: solving $\sin 3\theta = 0$ gives $3\theta = 0, \pi, 2\pi, \dots$, so $\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \dots$. The first petal runs from $\theta = 0$ to $\theta = \frac{\pi}{3}$, and

$$A_{\text{petal}} = \frac{1}{2}\int_0^{\pi/3} 4\sin^{2}3\theta\,d\theta = \frac{\pi}{3}.$$

Three petals give $\pi$ in total, which the integral over $[0, \pi]$ confirms directly.

Here is the trap that has no analogue in Unit 8. Integrating the same rose over $[0, 2\pi]$ gives

$$\frac{1}{2}\int_0^{2\pi} 4\sin^{2}3\theta\,d\theta = 2\pi,$$

which is exactly twice the rose's area of $\pi$. Nothing is wrong with the arithmetic. The three-petalled rose is completely traced by $[0, \pi]$, and the second half of the revolution redraws the same three petals, so the integral counts every sector twice.

The instinct being violated is a reasonable one: for a closed curve in rectangular coordinates, going all the way round is how you enclose the region. In polar form "all the way round" is a statement about the angle, and the curve may have finished long before the angle does.

Two habits that prevent it:

1. Never assume $[0, 2\pi]$. Determine the interval from the curve, by solving $r = 0$ or by tracking when the curve returns to its starting point.
2. Know the rose count. $r = a\sin(n\theta)$ has $n$ petals when $n$ is odd, traced over $[0, \pi]$, and $2n$ petals when $n$ is even, traced over $[0, 2\pi]$. The odd case is the one that double-counts.

This is 9.3's distinction in a new costume: an integral over too much parameter measures travel that the curve did not need. There the cost was a doubled arc length; here it is a doubled area.

Two further situations where the interval is not what it appears.

Negative $r$. Where $f(\theta) < 0$ the curve is plotted backwards along the ray, which places it in the opposite direction. The area integrand $r^{2}$ is unaffected by the sign, so a negative radius contributes area normally, and it is still the geometry that decides the limits. For $r = 2\sin 3\theta$ on $\left[\frac{\pi}{3}, \frac{2\pi}{3}\right]$ the radius is negative throughout, and that stretch traces a petal like any other.

Limaçons with an inner loop. A curve like $r = 1 + 2\cos\theta$ has an inner loop, drawn where $r < 0$. The outer loop and the inner loop need separate intervals, found the same way by solving $r = 0$, and the region between them is a difference rather than a single integral. Integrating over a full revolution adds the inner loop's area to the outer's instead of subtracting it.

The general statement worth carrying into 9.9: identify the interval from the geometry before writing any integral, and check whether the curve returns to the pole inside it. Every polar bounds error is a failure to do that first.