Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve · 求极坐标区域的面积或单条极坐标曲线围成的面积
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| area/ˈeərɪə/ | 面积 | miàn jī |
| sectors/ˈsektəz/ | 扇形 | shàn xíng |
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 $\theta$.
极曲线扫出的面积
- 对 $y=f(x)$ 我们把面积切成竖直矩形。极曲线则绕原点扫过。
- 所以我们把区域切成薄薄的饼状扇形,每个都是圆的一小片。
- 把这些扇形面积加起来,得到极曲线所围的面积。
- 结果是一个关于 $\theta$ 的干净积分。
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.
极坐标面积公式
- 角度 $\theta$ 处的薄扇形有半径 $r=f(\theta)$ 和微小角 $d\theta$;它的面积是 $\tfrac12 r^2\,d\theta$。
- 从 $\theta=\alpha$ 到 $\theta=\beta$ 相加:
-
$$A=\frac{1}{2}\int_{\alpha}^{\beta}\big(f(\theta)\big)^2\,d\theta$$
- 把半径平方,取一半,在角度区间上积分。
The area enclosed by $r=f(\theta)$ is... · $r=f(\theta)$所围成的面积为...
Half the integral of $r^2$. · $r^2$积分的一半。
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.
为何是 $\tfrac12 r^2$
- 半径 $r$、角 $\Delta\theta$ 的完整圆扇形面积是 $\tfrac12 r^2\,\Delta\theta$。
- 我们的薄片恰是这些扇形,$\Delta\theta\to d\theta$。
- 把它们求和(积分)累积扫出的区域。
- 这是"把薄矩形相加"的极坐标类比,改用饼片。
Thin sectors sweep the area · 薄扇形扫过该区域
A polar region is summed from thin pie-slice sectors, each of area $\tfrac12 r^2\,d\theta$. · 极坐标区域由薄扇形（每块面积为$\tfrac12 r^2\,d\theta$）累加而成。
The polar area formula sums the areas of thin circular ____. · 极坐标面积公式是对薄圆形____面积的求和。
Each sector has area $\tfrac12 r^2\,d\theta$. · 每个扇形的面积为$\tfrac12 r^2\,d\theta$。
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).
选择角度限
- 边界 $\alpha,\beta$ 是曲线把区域扫过一次的角度。
- 对简单曲线的一整圈,常是 $0$ 到 $2\pi$——但许多曲线更早闭合。
- 玫瑰线 $r=\cos(3\theta)$ 的一个花瓣在 $r$ 的相邻零点之间扫过。
- 选让区域恰好扫一次的边界(不重复计算)。
Find the area enclosed by $r=2$: $\tfrac12\int_0^{2\pi}4\,d\theta$ (as a multiple of $\pi$, enter the number). · 求$r=2$所围成的面积：$\tfrac12\int_0^{2\pi}4\,d\theta$ (以$\pi$的倍数形式填写数字)。
$\tfrac12\cdot4\cdot2\pi=4\pi$.
The polar area formula includes a factor of $\tfrac12$ and squares $r$. · 极坐标面积公式包含因子$\tfrac12$，并对$r$平方。
$A=\tfrac12\int r^2\,d\theta$.
The angle limits $\alpha,\beta$ should be chosen so the curve traces the region... · 角度极限$\alpha,\beta$的选择应使曲线恰好遍历该区域...
Tracing more than once double-counts area. · 重复遍历会导致面积重复计算。
The polar formula gives $4\pi$ for $r=2$, which matches... · 极坐标公式给出$4\pi$对于$r=2$，这与...
$\pi(2)^2=4\pi$ — the two agree. · $\pi(2)^2=4\pi$ — 两者一致。
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 $2\pi$).
公式是 $\tfrac12\int r^2\,d\theta$——别忘了 $\tfrac12$ 或 $r$ 上的平方。并选 $\alpha,\beta$ 使曲线恰好扫一次区域:角度范围超出会重复计算面积(例如一整个玫瑰花瓣需要两个相邻 $r=0$ 值之间的角度,而非 $0$ 到 $2\pi$)。
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 $\pi r^2=\pi(2)^2=4\pi$. ✓
求 $r=2$(半径 $2$ 的圆)所围的面积。
- $A=\dfrac12\displaystyle\int_0^{2\pi}(2)^2\,d\theta=\dfrac12\int_0^{2\pi}4\,d\theta=\dfrac12\cdot4\cdot2\pi=4\pi$。
- 这与熟悉的圆面积 $\pi r^2=\pi(2)^2=4\pi$ 一致。✓
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.
极曲线 $r=f(\theta)$ 所围的面积是 $A=\frac{1}{2}\int_{\alpha}^{\beta}\big(f(\theta)\big)^2\,d\theta$——把薄扇形 $\frac12 r^2\,d\theta$ 相加。保留 $\frac12$ 和平方,并选角度限使区域恰好扫一次。