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Area swept by a polar curve

• For $y=f(x)$ we sliced area into vertical rectangles. A polar curve sweeps around the origin instead.
• So we slice the region into thin pie-shaped sectors 扇形, each a sliver of a circle.
• Adding up those sector areas gives the area 面积 enclosed by the polar curve.
• The result is a clean integral in $\theta$.

The polar area formula

• A thin sector at angle $\theta$ has radius $r=f(\theta)$ and tiny angle $d\theta$; its area is $\tfrac12 r^2\,d\theta$.
• Add them from $\theta=\alpha$ to $\theta=\beta$:
• $$A=\frac{1}{2}\int_{\alpha}^{\beta}\big(f(\theta)\big)^2\,d\theta$$
• Square the radius, halve, and integrate over the angle interval.

The area swept by a polar curve between two angles is ½∫r² dθ

Why the $\tfrac12 r^2$

• The area of a full circular sector of radius $r$ and angle $\Delta\theta$ is $\tfrac12 r^2\,\Delta\theta$.
• Our thin slices are exactly these sectors with $\Delta\theta\to d\theta$.
• Summing them (integrating) accumulates the swept-out region.
• It's the polar analog of "sum thin rectangles," using pie slices instead.

Choosing the angle limits

• The limits $\alpha,\beta$ are the angles over which the curve sweeps the region once.
• For one full loop of a simple curve, that's often $0$ to $2\pi$ — but many curves close in less.
• A petal of a rose $r=\cos(3\theta)$ sweeps between consecutive zeros of $r$.
• Pick limits that trace the region exactly once (no double-counting).