Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
Introduced| English |
|---|
| area/ˈeərɪə/ |
| sectors/ˈsektəz/ |
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 $\theta=\beta$:
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$$A=\frac{1}{2}\int_{\alpha}^{\beta}\big(f(\theta)\big)^2\,d\theta$$
- Square the radius, halve, and integrate over the angle interval.
The area enclosed by $r=f(\theta)$ is...
Half the integral of $r^2$.
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.
Thin sectors sweep the area
A polar region is summed from thin pie-slice sectors, each of area $\tfrac12 r^2\,d\theta$.
The polar area formula sums the areas of thin circular ____.
Each sector has area $\tfrac12 r^2\,d\theta$.
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).
Find the area enclosed by $r=2$: $\tfrac12\int_0^{2\pi}4\,d\theta$ (as a multiple of $\pi$, enter the number).
$\tfrac12\cdot4\cdot2\pi=4\pi$.
The polar area formula includes a factor of $\tfrac12$ and squares $r$.
$A=\tfrac12\int r^2\,d\theta$.
The angle limits $\alpha,\beta$ should be chosen so the curve traces the region...
Tracing more than once double-counts area.
The polar formula gives $4\pi$ for $r=2$, which matches...
$\pi(2)^2=4\pi$ — the two agree.
The formula is $\tfrac12\int r^2\,d\theta$ — don't forget the $\tfrac12$ or the square on $r$. And choose $\alpha,\beta$ so the curve traces the region exactly once: overshooting the angle range double-counts area (a full rose petal, for instance, needs the angles between two consecutive $r=0$ values, not $0$ to $2\pi$).
Find the area enclosed by $r=2$ (a circle of radius $2$).
- $A=\dfrac12\displaystyle\int_0^{2\pi}(2)^2\,d\theta=\dfrac12\int_0^{2\pi}4\,d\theta=\dfrac12\cdot4\cdot2\pi=4\pi$.
- This matches the familiar circle area $\pi r^2=\pi(2)^2=4\pi$. ✓
The area enclosed by a polar curve $r=f(\theta)$ is $A=\frac{1}{2}\int_{\alpha}^{\beta}\big(f(\theta)\big)^2\,d\theta$ — summing thin sectors $\frac12 r^2\,d\theta$. Keep the $\frac12$ and the square, and choose the angle limits so the region is traced exactly once.