Polar regions between curves and inside both curves

Understand the idea

Polar area problems require identifying the allowed radial interval at each angle. For a region between two nonnegative radii, subtract the inner squared radius from the outer squared radius. For a region inside both curves and extending from the pole, the smaller radius determines the boundary.

Find intersection angles within the relevant domain and test which radius is larger on each subinterval. If the ordering changes, split the integral. Equating radius expressions at the same angle can miss geometric intersections when a curve uses negative radii or different parameter descriptions, so keep the sketch and the coordinate meaning in view.

The square of the difference of radii is not the difference of their squares. Decide first whether the problem asks for an overlap region, a region outside one curve but inside another, or a union. Those descriptions produce different radial boundaries.

A_{\text{between}}=\frac12\int(R^2-r^2)\,d\theta,\qquad A_{\text{common}}=\frac12\int\min(R^2,r^2)\,d\theta\quad(R,r\ge0)

A method you can apply

Translate the region description into radial inequalities on a chosen angle interval.
Locate every angle where the active inner or outer boundary changes.
Integrate the appropriate squared-radius expression on each subinterval and combine without double-counting.

Worked example

Find the first-quadrant area inside both $\displaystyle r=2$ and $\displaystyle r=4\cos\theta$ .

The angles range from $\displaystyle 0$ to $\displaystyle \pi/2$ . The radii agree when $\displaystyle 2=4\cos\theta$ , giving $\displaystyle \theta=\pi/3$ .
On $\displaystyle [0,\pi/3]$ , the smaller radius is $\displaystyle 2$ ; on $\displaystyle [\pi/3,\pi/2]$ , it is $\displaystyle 4\cos\theta$ .
Thus $\displaystyle A=\tfrac12\int_0^{\pi/3}4\,d\theta+\tfrac12\int_{\pi/3}^{\pi/2}16\cos^2\theta\,d\theta$ .
The first term is $\displaystyle 2\pi/3$ . The second is $\displaystyle 2\pi/3-\sqrt3$ after using the double-angle identity.

The common area in the first quadrant is

\displaystyle 4\pi/3-\sqrt3

.

Try it yourself

Find the upper-half-plane area inside $\displaystyle r=3$ and outside $\displaystyle r=1$ .

Show solution and reasoning

For $\displaystyle 0\le\theta\le\pi$ , the outer and inner radii are $\displaystyle 3$ and $\displaystyle 1$ . The area is $\displaystyle \tfrac12\int_0^\pi(9-1)\,d\theta=4\pi$ .