Wemazi

Topic 9.8 · BC

Calculate the requested area

Builds on setting up the polar area integral. 8 minutes.

Two equal halves

Because \(\cos(-\theta) = \cos\theta\), the cardioid \(r = 1 + \cos\theta\) is symmetric about the x-axis. Its top half, for \(0 \le \theta \le \pi\), is the mirror image of its bottom half.

Press the button to reflect the top half in the x-axis. The copy covers the bottom half, so the two halves have equal area.

The whole area is twice the area of the top half. The factor 2 cancels the \(\frac12\) in the area formula, so the area is \(\int_0^{\pi}(1 + \cos\theta)^2\,d\theta\).

Expand the square to \(1 + 2\cos\theta + \cos^2\theta\). The three terms integrate to \(\pi\), 0 and \(\frac{\pi}{2}\), so the area is \(\frac{3\pi}{2}\).

Equal pieces

A region may be made of equal pieces. The two halves of a cardioid are mirror images, and the petals of a rose can turn onto each other.

Press the button that turns the teal petal onto the next petal. The copy covers the next petal, so the two petals have equal area.

Multiply the area of one piece by the number of pieces, and count the pieces in the figure. The equation can mislead, since \(r = 4\cos 2\theta\) has four petals, not two.

Whether the pieces fit is a fact about the region, so check it before you multiply.

Notation

  1. Integrate half the squared radius over an interval that traces the region once.

    \[\cb{A} = \int_{\alpha}^{\beta} \tfrac12\,[f(\theta)]^2\,d\theta\]

  2. If a reflection or a rotation maps each of \(k\) pieces onto the next, integrate over one piece and multiply by \(k\).

    \[\cb{A} = k \cdot \int_{\text{one piece}} \tfrac12\,[f(\theta)]^2\,d\theta\]

  3. If a substitution leaves \(f(\theta)\) unchanged, the curve is symmetric about that axis.

    \[\begin{aligned} f(-\theta) &= f(\theta) && \text{x-axis} \\ f(\pi - \theta) &= f(\theta) && \text{y-axis} \end{aligned}\]

  4. Use the half-angle identities to integrate squares of sine and cosine.

    \[\begin{aligned} \cos^2 u &= \tfrac{1 + \cos 2u}{2} \\ \sin^2 u &= \tfrac{1 - \cos 2u}{2} \end{aligned}\]

  5. Use symmetry once. If the interval already covers the whole region, do not multiply again.

When it fails

Slide \(c\) to 2, which gives \(r = 3 - 2\sin\theta\), and reflect the top half. Because \(f(-\theta) = 3 + 2\sin\theta\), the curve is not symmetric about the x-axis, and the copy covers only part of the bottom half.

The top half has area about 5.3, and the bottom half about 29.3. Doubling the top half gives about 10.6, but the whole area is \(11\pi \approx 34.6\).

Example 1: a rose with three petals

Find the area inside \(r = 4\sin 3\theta\).

  1. Because \(r(\theta + \pi) = -r(\theta)\), the interval \(0 \le \theta \le \pi\) traces the three petals once. One petal runs between neighboring zeros of \(r\), such as 0 and \(\frac{\pi}{3}\). Find its area with a half-angle identity.

    \[\begin{aligned} A_{\text{petal}} &= \tfrac12\int_0^{\pi/3} 16\sin^2 3\theta\,d\theta \\ &= 8\int_0^{\pi/3} \tfrac{1 - \cos 6\theta}{2}\,d\theta \end{aligned}\]

  2. Integrate.

    \[A_{\text{petal}} = 4\left[\theta - \tfrac{\sin 6\theta}{6}\right]_0^{\pi/3} = \tfrac{4\pi}{3}\]

  3. A rotation by \(\frac{2\pi}{3}\) turns each petal onto the next, so the three petals are equal. Multiply by 3.

    \[A = 3\cdot\tfrac{4\pi}{3} = 4\pi\]

  4. As a check, integrate over the whole interval without symmetry. The half-angle identity gives \(\frac{\pi}{2}\) for the integral of \(\sin^2 3\theta\).

    \[\begin{aligned} A &= \tfrac12\int_0^{\pi} 16\sin^2 3\theta\,d\theta \\ &= 8\cdot\tfrac{\pi}{2} = 4\pi \end{aligned}\]

Example 2: using symmetry correctly

Find the area inside \(r = 4 - 2\sin\theta\).

  1. The radius is at least 2, so it is never negative. One full turn, \(0 \le \theta \le 2\pi\), traces the region once.

  2. A tempting wrong move: The region looks symmetric, so multiply the whole-turn integral by 2.

    \[\begin{aligned} A &= 2\cdot\tfrac12\int_0^{2\pi}(4 - 2\sin\theta)^2\,d\theta \\ &= 36\pi \end{aligned}\]

    The curve is never more than 6 units from the pole, so the region lies inside the red disk of area \(36\pi\). The curve touches the edge of that disk at one point only, so the area is less than \(36\pi\).

  3. Symmetry replaces part of the turn, not the whole turn. Replacing \(\theta\) with \(\pi - \theta\) leaves \(\sin\theta\) unchanged, so the curve is symmetric about the y-axis. Integrate over the right half, \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\), and double it.

    \[\begin{aligned} A &= 2\cdot\tfrac12\int_{-\pi/2}^{\pi/2}(4 - 2\sin\theta)^2\,d\theta \\ &= \int_{-\pi/2}^{\pi/2}(16 - 16\sin\theta \\ &\qquad + 4\sin^2\theta)\,d\theta \end{aligned}\]

  4. The three terms integrate to \(16\pi\), 0 and \(4\cdot\frac{\pi}{2}\).

    \[A = 16\pi - 0 + 2\pi = 18\pi\]

  5. The wrong value \(36\pi\) is twice \(18\pi\), because it counted the whole region twice.

Practice

Find the area inside the rose \(r = 2\sin 2\theta\).

Decide the interval first: compare \(f(\theta + \pi)\) with \(f(\theta)\). Then use \(\sin^2 u = \frac{1 - \cos 2u}{2}\).

Solution

  1. The radius has the same sign after a half turn, so the second half turn draws two new petals. The rose needs a full turn, \(0 \le \theta \le 2\pi\).

    \[r(\theta + \pi) = r(\theta)\]

  2. Set up the integral and use the half-angle identity.

    \[\begin{aligned} A &= \tfrac12\int_0^{2\pi} 4\sin^2 2\theta\,d\theta \\ &= \int_0^{2\pi}(1 - \cos 4\theta)\,d\theta \end{aligned}\]

  3. Integrate.

    \[A = \left[\theta - \tfrac{\sin 4\theta}{4}\right]_0^{2\pi} = 2\pi\]

  4. As a check, one petal (\(0 \le \theta \le \frac{\pi}{2}\)) has area \(\frac{\pi}{2}\). Four equal petals give \(4\cdot\frac{\pi}{2} = 2\pi\).

Questions

The figure shows the rose \(r = 6\cos 2\theta\), which has four equal petals. The teal petal has area \(\frac{9\pi}{2}\).

  1. One petal has area \(\frac{9\pi}{2}\). What is the area inside the whole rose?

    • \(18\pi\)
    • \(9\pi\)
    • \(36\pi\)
    Answer

    \(18\pi\)

  2. Which integral gives the area inside the whole rose?

    • \(\displaystyle 4\cdot\tfrac12\int_{-\pi/4}^{\pi/4}(6\cos 2\theta)^2\,d\theta\)
    • \(\displaystyle 2\cdot\tfrac12\int_{-\pi/4}^{\pi/4}(6\cos 2\theta)^2\,d\theta\)
    • \(\displaystyle \tfrac12\int_0^{\pi}(6\cos 2\theta)^2\,d\theta\)
    • \(\displaystyle 2\cdot\tfrac12\int_0^{2\pi}(6\cos 2\theta)^2\,d\theta\)
    Answer

    \(\displaystyle 4\cdot\tfrac12\int_{-\pi/4}^{\pi/4}(6\cos 2\theta)^2\,d\theta\)

  3. What is the area inside \(r = 4 + \cos\theta\)?

    • \(\frac{33\pi}{2}\)
    • \(16\pi\)
    • \(33\pi\)
    • \(17\pi\)
    Answer

    \(\frac{33\pi}{2}\)

Summary

The area inside a polar curve is the integral of half the squared radius over angles that trace the region once.

If a reflection or rotation maps each piece onto the next, the pieces are equal, so you multiply one piece's area by their number.

You count the pieces in the picture, and you never multiply an integral that already covers the whole region.

This is one of four skills in topic 9.8 of Which & Why: AP Calculus.