Area swept by one polar curve | Code Scholars Tutoring

Understand the idea

A small polar sector has area approximately one half of radius squared times its angle width. Accumulating these sectors gives the polar area formula. Correct bounds must describe the intended region with the intended number of traversals.

Sketch or analyze the curve before selecting an angle interval. Zeros of the radius, symmetry, and key angle values help determine where a loop or petal begins and ends. A familiar full-turn interval can trace some polar curves more than once and therefore count an area repeatedly.

The squared radius makes the sector contribution nonnegative, but it does not determine whether the same physical region has already been counted. When using symmetry, identify a specific fraction of the region and justify the multiplier.

A=\frac12\int_{\alpha}^{\beta}[r(\theta)]^2\,d\theta

A method you can apply

Identify the intended region and an angle interval that traces its boundary appropriately.
Square the complete radius expression and include the factor $\displaystyle 1/2$ .
Integrate, using identities or symmetry when justified, and check for repeated tracing.

Worked example

Find the area enclosed by $\displaystyle r=2\sin\theta$ .

The radius is nonnegative on $\displaystyle 0\le\theta\le\pi$ , and this interval traces the circle once.
Set up $\displaystyle A=\tfrac12\int_0^\pi4\sin^2\theta\,d\theta$ .
Using $\displaystyle \sin^2\theta=(1-\cos2\theta)/2$ , the integral becomes $\displaystyle \int_0^\pi(1-\cos2\theta)\,d\theta$ .

The area is

\displaystyle \pi

, consistent with a circle of radius

\displaystyle 1

centered at

\displaystyle (0,1)

.

Try it yourself

Find the sector area swept by $\displaystyle r=3$ for $\displaystyle 0\le\theta\le\pi/3$ .

Show solution and reasoning

The area is $\displaystyle \tfrac12\int_0^{\pi/3}9\,d\theta=3\pi/2$ . This is one sixth of the area of the circle of radius $\displaystyle 3$ .