Which integral gives the area of the teal region?

\int_0^{\pi/3} \tfrac12(4\cos\theta)^2\,d\theta

\int_0^{\pi/3} \tfrac12(4\cos\theta)\,d\theta

\int_0^{\pi/3} (4\cos\theta)^2\,d\theta

\int_0^{\pi/6} \tfrac12(4\cos\theta)^2\,d\theta

Which integral gives the area inside r = 5 - 2\cos\theta?

\tfrac12\int_0^{2\pi} (5 - 2\cos\theta)^2\,d\theta

\tfrac12\int_0^{2\pi} (25 - 4\cos^2\theta)\,d\theta

\tfrac12\int_0^{\pi} (5 - 2\cos\theta)^2\,d\theta

\int_0^{2\pi} (5 - 2\cos\theta)^2\,d\theta

Topic 9.8 · BC

Set up the polar area integral

Builds on checking whether a polar curve repeats. 7 minutes.

Slices that are almost sectors

The spiral \(r = \theta\) winds outward from the pole. The region lies between the pole, the spiral and the rays \(\theta = 0\) and \(\theta = 2\). It is not a circular sector, because the radius grows.

Cut the region into slices with rays of equal angle. Replace each slice with a teal circular sector whose radius is the spiral's radius at the slice's first ray.

A sector with radius \(r\) and angle \(\Delta\theta\) is the fraction \(\frac{\Delta\theta}{2\pi}\) of a disk of area \(\pi r^2\). Its area is \(\frac{\Delta\theta}{2\pi}\cdot\pi r^2 = \frac12 r^2\,\Delta\theta\).

With 4 slices of angle 0.5, the radii are 0, 0.5, 1 and 1.5. The sector areas add up to \(\frac12(0^2 + 0.5^2 + 1^2 + 1.5^2)(0.5) = 0.875\).

Drag the slider to use more slices. The sectors fit the region better, and the total grows toward the area of the region, \(\frac43 \approx 1.333\).

A sum of thin sectors

The curve is now a general \(r = f(\theta)\), and the region lies between the rays \(\theta = \alpha\) and \(\theta = \beta\).

The slice outlined in orange has the small angle \(\cy{d\theta}\), and its radius \(\cy{r}\) is about \(f(\theta)\). The slice is almost a circular sector, so its area is about \(\frac12 f(\theta)^2\,d\theta\).

Drag the first slider to make the slices thinner. The radius changes less across a thin slice, so the error in each slice is smaller.

As the slices become thinner, the sum of the sector areas approaches the area of the region. This limit is an integral whose integrand is half the squared radius and whose variable is the angle.

Notation

A thin slice with the angle \(\cy{d\theta}\) and the radius \(\cy{r}\) has about the area of a circular sector.
\[\cb{dA} = \tfrac12\,\cy{r}^2\,\cy{d\theta}\]
Add all the slices from the first ray to the last ray.
\[\cb{A} = \int_{\alpha}^{\beta} \tfrac12\,[f(\theta)]^2\,d\theta\]
The angles from \(\alpha\) to \(\beta\) must trace the region once, so check on the graph that the curve does not repeat. The function \(f\) must be continuous on \([\alpha, \beta]\).
To set up the integral, write \(r\) as a function of \(\theta\), square the whole expression and multiply by \(\frac12\).
Square the whole radius, not each term, because the square of a sum has a middle term.
\[(2 + \sin\theta)^2 = 4 + 4\sin\theta + \sin^2\theta\]
The squared radius is never negative, so a negative \(r\) adds area, as a positive \(r\) does.

Example 1: a sector of a circle

Set up the integral for the area of the part of the disk \(r = 4\) between the rays \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{2\pi}{3}\). Check it with geometry.

The radius is constant, so \(f(\theta) = 4\). The rays are \(\alpha = \frac{\pi}{6}\) and \(\beta = \frac{2\pi}{3}\).
Set up the integral.
\[A = \int_{\pi/6}^{2\pi/3} \tfrac12\,(4)^2\,d\theta\]
The integrand is the constant 8, so the integral is 8 times the angle.
\[A = 8\left(\tfrac{2\pi}{3} - \tfrac{\pi}{6}\right) = 8\cdot\tfrac{\pi}{2} = 4\pi\]
Check with geometry. The angle \(\frac{\pi}{2}\) is a quarter turn, and a quarter of a disk of radius 4 has area \(\frac14\cdot 16\pi = 4\pi\).

Example 2: a radius that is a sum

Set up the integral for the area inside \(r = 3 + \cos 2\theta\). Then check that the value is reasonable.

The radius is at least 2, so it is positive at every angle and one full turn traces the region once.
\[0 \le \theta \le 2\pi\]
A tempting wrong move: Integrate the radius, not its square.
\[\begin{aligned} A &= \int_0^{2\pi} \tfrac12\,(3 + \cos 2\theta)\,d\theta \\ &= 3\pi \end{aligned}\]
The curve stays at least 2 units from the pole, so the region contains the disk inside the red circle of radius 2. That disk alone has area \(4\pi\), which is more than \(3\pi\).
A thin slice has area \(\frac12 r^2\,d\theta\). Square the whole radius.
\[A = \int_0^{2\pi} \tfrac12\,(3 + \cos 2\theta)^2\,d\theta\]
The square is \(9 + 6\cos 2\theta + \cos^2 2\theta\), and over a full turn its terms integrate to \(18\pi\), 0 and \(\pi\). The radius is between 2 and 4, so the area must lie between \(\pi\cdot 2^2 = 4\pi\) and \(\pi\cdot 4^2 = 16\pi\), and \(\frac{19\pi}{2}\) does.
\[\begin{aligned} A &= \tfrac12\left(18\pi + 0 + \pi\right) \\ &= \tfrac{19\pi}{2} \approx 29.8 \end{aligned}\]

Practice

Set up and evaluate the area of the region between the pole, the curve \(r = \sqrt{\theta}\), and the rays \(\theta = 0\) and \(\theta = \pi\).

The orange line has length \(\sqrt{\theta}\). Square it before you integrate, so the square root disappears.

Solution

Square the radius: \(r^2 = \theta\). Set up the integral.
\[A = \int_0^{\pi} \tfrac12\,\theta\,d\theta\]
Integrate and substitute the bounds.
\[A = \tfrac12\cdot\tfrac{\theta^2}{2}\Big|_0^{\pi} = \tfrac{\pi^2}{4}\]
As a check, \(\sqrt{\theta}\) is at most \(\sqrt{\pi}\), so the region fits inside the faint half circle of radius \(\sqrt{\pi}\). That half disk has area \(\frac{\pi^2}{2} \approx 4.93\), and \(\frac{\pi^2}{4} \approx 2.47\) is smaller, as it must be.

Questions

The figure shows the circle \(r = 4\cos\theta\) and the teal region inside it between the rays at 0 and \(\frac{\pi}{3}\). The sector outlined in orange is a thin slice with radius 3 and angle 0.2.

About how large is the area of the thin slice that is outlined in orange?
- 0.9
- 0.6
- 1.8
- 0.3
Answer
0.9
Which integral gives the area of the teal region?
- \(\displaystyle\int_0^{\pi/3} \tfrac12(4\cos\theta)^2\,d\theta\)
- \(\displaystyle\int_0^{\pi/3} \tfrac12(4\cos\theta)\,d\theta\)
- \(\displaystyle\int_0^{\pi/3} (4\cos\theta)^2\,d\theta\)
- \(\displaystyle\int_0^{\pi/6} \tfrac12(4\cos\theta)^2\,d\theta\)
Answer
\(\displaystyle\int_0^{\pi/3} \tfrac12(4\cos\theta)^2\,d\theta\)
Which integral gives the area inside \(r = 5 - 2\cos\theta\)?
- \(\displaystyle\tfrac12\int_0^{2\pi} (5 - 2\cos\theta)^2\,d\theta\)
- \(\displaystyle\tfrac12\int_0^{2\pi} (25 - 4\cos^2\theta)\,d\theta\)
- \(\displaystyle\tfrac12\int_0^{\pi} (5 - 2\cos\theta)^2\,d\theta\)
- \(\displaystyle\int_0^{2\pi} (5 - 2\cos\theta)^2\,d\theta\)
Answer
\(\displaystyle\tfrac12\int_0^{2\pi} (5 - 2\cos\theta)^2\,d\theta\)

Summary

You set up the area as \(\frac12\int_{\alpha}^{\beta} [f(\theta)]^2\,d\theta\), with the whole radius squared and with angles that trace the region once.

This works because a thin slice is almost a circular sector, and a sector with radius \(r\) and angle \(d\theta\) has area \(\frac12 r^2\,d\theta\).

You use it for the area of a region between the pole and a polar curve.

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