Area of a Polar Region

Areas between graphs came from thin rectangles. For a polar curve, the natural pieces are thin sectors, wedges of a circle with their point at the origin.

The area formula

A sector of a circle of radius

r

with angle

\Delta\theta

has area

\frac{1}{2}r^2\Delta\theta

. Cut the region swept out by

r = f(\theta)

into thin sectors, each with radius

f(\theta)

at some angle in its piece. Adding them gives a Riemann sum.

Four thin sectors under r = 5 + 3 cos θ. As Δθ shrinks, their total area approaches the area of the region.

The limits are angles. They come from the rays that bound the region, or from the angles where

r = 0

, where the curve passes through the origin. For a whole closed curve, they're one full trip around it. Because

r

is squared, the formula also works where

r < 0

Example 1

Find the area of the region enclosed by

r = 2 + \cos 2\theta

Since

r > 0

for every

\theta

, the curve goes around the origin once as

\theta

runs from 0 to

2\pi

\begin{aligned} A &= \frac{1}{2}\int_{0}^{2\pi} (2 + \cos 2\theta)^2\,d\theta \\[4pt] &= \frac{1}{2}\int_{0}^{2\pi} \big(4 + 4\cos 2\theta \\ &\qquad\qquad + \cos^2 2\theta\big)\,d\theta. \end{aligned}

Over

[0, 2\pi]

the integral of

\cos 2\theta

is 0. By the identity

\cos^2 u = \frac{1}{2}(1 + \cos 2u)

, the last term integrates to half the length of the interval,

\pi

A = \frac{1}{2}(8\pi + 0 + \pi) = \frac{9\pi}{2} \approx 14.137.

The region enclosed by r = 2 + cos 2θ.

An inner loop

Example 2

The curve

r = 1 + \sqrt{2}\cos\theta

has a small inner loop. Find the area enclosed by the inner loop.

The curve passes through the origin where

\cos\theta = -\frac{1}{\sqrt{2}}

, at

\theta = \frac{3\pi}{4}

and

\theta = \frac{5\pi}{4}

. Between them

r < 0

, and the point traces the inner loop once.

\begin{aligned} A &= \frac{1}{2}\int_{3\pi/4}^{5\pi/4} \left(1 + \sqrt{2}\cos\theta\right)^2 d\theta \\[4pt] &= \frac{1}{2}\int_{3\pi/4}^{5\pi/4} \big(1 + 2\sqrt{2}\cos\theta \\ &\qquad\qquad + 2\cos^2\theta\big)\,d\theta. \end{aligned}

With

2\cos^2\theta = 1 + \cos 2\theta

, an antiderivative of the integrand is

2\theta + 2\sqrt{2}\sin\theta + \frac{1}{2}\sin 2\theta

. Its change from

\frac{3\pi}{4}

\frac{5\pi}{4}

\begin{aligned} &\pi + 2\sqrt{2}\left(-\sqrt{2}\right) + \tfrac{1}{2}\big(1 - (-1)\big) \\ &\qquad = \pi - 4 + 1 = \pi - 3. \end{aligned}

So the area is

\frac{\pi - 3}{2} \approx 0.071

The inner loop, shaded, traced as θ runs from 3π/4 to 5π/4. On those angles r is negative, so the loop sits opposite the dashed rays.

A region between two rays

Example 3

Find the area of the part of the region enclosed by

r = 5 + 3\cos\theta

that lies in the second quadrant.

The second quadrant is swept as

\theta

runs from

\frac{\pi}{2}

\pi

, and

r > 0

there. The double-angle identity rewrites the squared cosine as

9\cos^2\theta = \frac{9}{2} + \frac{9}{2}\cos 2\theta

\begin{aligned} A &= \frac{1}{2}\int_{\pi/2}^{\pi} (5 + 3\cos\theta)^2\,d\theta \\[4pt] &= \frac{1}{2}\int_{\pi/2}^{\pi} \big(\tfrac{59}{2} + 30\cos\theta \\ &\qquad\qquad + \tfrac{9}{2}\cos 2\theta\big)\,d\theta \\[4pt] &= \frac{1}{2}\Big(\tfrac{59}{2}\theta + 30\sin\theta \\ &\qquad\qquad + \tfrac{9}{4}\sin 2\theta\Big)\Big|_{\pi/2}^{\pi} \\[4pt] &= \frac{1}{2}\left(\frac{59\pi}{2} - \frac{59\pi}{4} - 30\right) \\[4pt] &= \frac{59\pi}{8} - 15 \approx 8.169. \end{aligned}

The part of the region inside r = 5 + 3 cos θ in the second quadrant. It lies between the rays θ = π/2 and θ = π.