Wemazi

Topic 9.9 · BC

Split the angle interval when needed

Builds on naming the outer and inner curves on each stretch. 8 minutes.

An edge that changes curves

The solid curve is \(r = 3 + 2\cos\theta\), and the dashed curve is the circle \(r = 4\). The teal region is their common interior, the region inside both curves.

On each ray, this region runs from the pole to the nearer curve, so the nearer curve is its edge, drawn in orange.

Turn the ray to \(\theta = 0\). The solid curve has \(r = 5\) and the circle has \(r = 4\), so the edge is on the circle.

Turn the ray to \(\theta = \pi\). The solid curve has \(r = 1\), so the edge is on the solid curve.

The edge changes curves at \(\theta = \frac{\pi}{3}\), where \(3 + 2\cos\theta = 4\), so one formula cannot describe the edge on both sides.

One region, two formulas

Drag the angle across a crossing. The edge of the common interior follows one curve up to the crossing and the other curve after it.

Press Show the region between the curves. On each ray, this region runs from the nearer curve, \(\cy{r_1}\), to the farther curve, \(\cy{r_2}\), and the labels swap curves at the same crossing.

In both cases, the formula changes at a crossing. Between two crossings, the same curves play the same roles, so each stretch needs its own integral.

Notation

  1. Find an angle \(c\) between \(a\) and \(b\) where the curves meet. It cuts the range into two stretches.

    \[a < c < b\]

  2. On each stretch, decide which curve is outer or, for a common interior, which curve is nearer.

  3. For a region between two curves, write one integral for each stretch. Add the integrals.

    \[\begin{aligned} A &= \cb{\tfrac12\int_a^{c} \big(r_2^2 - r_1^2\big)\,d\theta} \\ &\quad + \cb{\tfrac12\int_c^{b} \big(r_2^2 - r_1^2\big)\,d\theta} \end{aligned}\]

    The curves called \(r_1\) and \(r_2\) can be different on the two stretches.

  4. For a common interior, integrate half the square of the nearer curve on each stretch.

    \[A = \cb{\tfrac12\int_a^{c} r_A^2\,d\theta} + \cb{\tfrac12\int_c^{b} r_B^2\,d\theta}\]

    Here \(r_A\) and \(r_B\) are the nearer curves on the first and second stretch.

  5. One integral over the whole range is correct only when the same curves play the same roles throughout.

When it fails

Suppose you use the circle \(r = 4\) as the edge for every angle. From \(\theta = \frac{\pi}{3}\) to \(\theta = \frac{5\pi}{3}\), the circle is beyond the solid curve.

The red strip lies inside the circle but outside the solid curve, so it is not in the common interior. The integral counts it anyway, so the result is too large.

Example 1: two circles

Find the area of the region inside both \(r = 6\cos\theta\) and \(r = 3\).

  1. The curves meet where \(6\cos\theta = 3\). Then \(\cos\theta = \tfrac12\), so \(\theta = \pm\frac{\pi}{3}\).

    \[6\cos\theta = 3\]

  2. At \(\theta = 0\), the values are 6 and 3, so the circle is nearer. For \(0 \le \theta \le \frac{\pi}{3}\), the edge is \(r = 3\).

    \[\cy{r_A = 3}\]

  3. After \(\frac{\pi}{3}\), \(6\cos\theta\) is less than 3, so the edge is \(r = 6\cos\theta\) up to \(\frac{\pi}{2}\), where that curve reaches the pole.

    \[\cy{r_B = 6\cos\theta}\]

  4. The region is symmetric about the x-axis. Find the upper half and double it.

    \[\begin{aligned} A = 2\Big[&\tfrac12\int_0^{\pi/3} 9\,d\theta \\ &+ \tfrac12\int_{\pi/3}^{\pi/2} 36\cos^2\theta\,d\theta\Big] \end{aligned}\]

  5. For the second integral, use \(\cos^2\theta = \frac{1 + \cos 2\theta}{2}\). The two integrals are \(\frac{3\pi}{2}\) and \(\frac{3\pi}{2} - \frac{9\sqrt3}{4}\).

    \[\begin{aligned} \tfrac12\int_0^{\pi/3} 9\,d\theta &= \tfrac{3\pi}{2} \\ \tfrac12\int_{\pi/3}^{\pi/2} 36\cos^2\theta\,d\theta &= \tfrac{3\pi}{2} - \tfrac{9\sqrt3}{4} \end{aligned}\]

  6. Add the two integrals and double the sum.

    \[\cb{A = 6\pi - \tfrac{9\sqrt3}{2}}\]

Example 2: two curves that swap

Find the area of the region between \(r = 2 + \cos\theta\) and \(r = 2 + \sin\theta\), for \(0 \le \theta \le \frac{\pi}{2}\).

  1. The curves meet where \(\cos\theta = \sin\theta\), so \(\theta = \frac{\pi}{4}\). The first curve is outer before \(\frac{\pi}{4}\), and the second is outer after it.

    \[2 + \cos\theta = 2 + \sin\theta \;\Rightarrow\; \theta = \tfrac{\pi}{4}\]

  2. A tempting wrong move: Use the first curve as the outer curve for the whole range.

    \[\begin{aligned} A &= \tfrac12\int_0^{\pi/2}\big[(2 + \cos\theta)^2 \\ &\qquad - (2 + \sin\theta)^2\big]\,d\theta \\ &= 0 \end{aligned}\]

    The graph shows a region with positive area. On the second stretch, the first curve is inner, so the integral counts that part, in red, as negative, and it cancels the teal part.

  3. Split at \(\frac{\pi}{4}\). Let \(D\) be the squared first radius minus the squared second radius. On the second stretch, the outer radius is the second curve, so use \(-D\).

    \[\begin{aligned} D &= (2+\cos\theta)^2 - (2+\sin\theta)^2 \\ A &= \tfrac12\int_0^{\pi/4} D\,d\theta \\ &\quad + \tfrac12\int_{\pi/4}^{\pi/2} (-D)\,d\theta \end{aligned}\]

  4. \(D\) expands to \(4(\cos\theta - \sin\theta) + \cos 2\theta\), because \(\cos^2\theta - \sin^2\theta = \cos 2\theta\). The mirror image of the first stretch in the ray \(\theta = \frac{\pi}{4}\) fills the second, so the two integrals are equal.

    \[\begin{aligned} \tfrac12\int_0^{\pi/4}\big[&4(\cos\theta - \sin\theta) \\ &+ \cos 2\theta\big]d\theta = 2\sqrt2 - \tfrac74 \end{aligned}\]

  5. Add the two equal integrals.

    \[\cb{A = 4\sqrt2 - \tfrac72}\]

Practice

Find the area of the region inside both \(r = 1 + \sin\theta\) and \(r = \tfrac32\).

Find where the curves meet. Then turn the ray to one angle in each stretch and see which dot is nearer the pole.

Solution

  1. The curves meet where \(1 + \sin\theta = \frac32\). Then \(\sin\theta = \frac12\), so \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

    \[1 + \sin\theta = \tfrac32\]

  2. At \(\theta = \frac{\pi}{2}\), the values are 2 and \(\frac32\), so the circle is the edge from \(\frac{\pi}{6}\) to \(\frac{5\pi}{6}\). At \(\theta = \frac{3\pi}{2}\), they are 0 and \(\frac32\), so \(r = 1 + \sin\theta\) is the edge for the rest of the turn, from \(\frac{5\pi}{6}\) to \(\frac{13\pi}{6} = \frac{\pi}{6} + 2\pi\).

  3. Write one integral for each stretch.

    \[\begin{aligned} A &= \tfrac12\int_{\pi/6}^{5\pi/6}\tfrac94\,d\theta \\ &\quad + \tfrac12\int_{5\pi/6}^{13\pi/6}(1 + \sin\theta)^2\,d\theta \end{aligned}\]

  4. For the second integral, expand the square and use \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\). The integrals are \(\frac{3\pi}{4}\) and \(\pi - \frac{9\sqrt3}{8}\).

    \[A = \tfrac{3\pi}{4} + \pi - \tfrac{9\sqrt3}{8} = \cb{\tfrac{7\pi}{4} - \tfrac{9\sqrt3}{8}}\]

Questions

Questions 1 and 2 use the region inside both the curve \(r = 2 + 2\sin\theta\) and the circle \(r = 3\).

  1. At \(\theta = \frac{\pi}{2}\), which curve forms the edge of the shaded region?

    • The circle \(r = 3\)
    • The curve \(r = 2 + 2\sin\theta\)
    • Neither, because the curves meet there.
    Answer

    The circle \(r = 3\)

  2. For which angles in \([0, \pi]\) is the circle the edge?

    • \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\)
    • \(0 \le \theta \le \frac{\pi}{6}\) and \(\frac{5\pi}{6} \le \theta \le \pi\)
    • \(0 \le \theta \le \pi\)
    Answer

    \(\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}\)

  3. The circle is the edge from \(\frac{\pi}{6}\) to \(\frac{5\pi}{6}\). What is the area of that part of the region, \(\tfrac12\int_{\pi/6}^{5\pi/6} 9\,d\theta\)?

    • \(3\pi\)
    • \(6\pi\)
    • \(\tfrac{9\pi}{2}\)
    Answer

    \(3\pi\)

Summary

You find the angles where the curves meet and cut the angle range at each one.

Between two crossings, the same curves bound the region, so one integral covers each stretch and the integrals add to the area.

You split the range whenever the outer curve, the inner curve or the edge of a common interior changes at a crossing.

This is one of four skills in topic 9.9 of Which & Why: AP Calculus.