Finding the Area of the Region Bounded by Two Polar Curves · 求两条极坐标曲线围成区域的面积
The region between two polar curves
- Just as with $x,y$ curves, you can find the area between two polar curves.
- Take the area inside the outer curve and subtract the area inside the inner curve.
- Each is a $\tfrac12\int r^2\,d\theta$ integral; the answer is their difference.
- The trick is finding where the curves cross to set the angle limits.
两条极曲线之间的区域
- 就像 $x,y$ 曲线一样,你可以求两条极曲线之间的面积。
- 取外曲线内的面积,减去内曲线内的面积。
- 每个都是 $\tfrac12\int r^2\,d\theta$ 积分;答案是它们的差。
- 诀窍是找到曲线相交处以确定角度限。
Find the intersection angles first
- Set the two curves equal, $f(\theta)=g(\theta)$, and solve for the intersection angles.
- These angles are the limits of integration $\alpha,\beta$ for the enclosed region.
- (Polar intersections can be sneaky — a point may be reached at different $\theta$; sketch to be sure.)
- The region of interest lies between these angles.
先找交点角度
- 令两曲线相等 $f(\theta)=g(\theta)$,解出交点角度。
- 这些角度就是围成区域的积分限 $\alpha,\beta$。
- (极坐标交点可能狡猾——一个点可能在不同 $\theta$ 处到达;画图确认。)
- 感兴趣的区域位于这些角度之间。
The area between two polar curves is $\tfrac12\int_\alpha^\beta(\ ?\ )\,d\theta$ where the integrand is... · 两条极坐标曲线间的面积为$\tfrac12\int_\alpha^\beta(\ ?\ )\,d\theta$，其中被积函数为...
Outer squared minus inner squared. · 外曲线平方减去内曲线平方。
To set the angle limits, find the ____ angles by solving $f(\theta)=g(\theta)$. · 为确定角度极限，需通过解$f(\theta)=g(\theta)$求出____角。
Where the curves cross bounds the region. · 曲线相交处界定了该区域。
Outer squared minus inner squared
- On the interval, identify which curve is outer (larger $r$) and which is inner.
- The area between them is:
-
$$A=\frac{1}{2}\int_{\alpha}^{\beta}\Big(\big[r_{\text{outer}}\big]^2-\big[r_{\text{inner}}\big]^2\Big)\,d\theta$$
- Square each radius, subtract, halve, integrate — the polar washer.
外平方减内平方
- 在区间上,判断哪条曲线是外($r$ 更大)、哪条是内。
- 它们之间的面积是:
-
$$A=\frac{1}{2}\int_{\alpha}^{\beta}\Big(\big[r_{\text{outer}}\big]^2-\big[r_{\text{inner}}\big]^2\Big)\,d\theta$$
- 把每个半径平方,相减,取一半,积分——极坐标垫圈。
Between two polar curves · 两条极坐标曲线之间
The area between two polar curves subtracts the inner sectors from the outer — $\tfrac12\int(r_\text{out}^2-r_\text{in}^2)\,d\theta$. · 两条极坐标曲线间的面积等于外曲线扇形减去内曲线扇形 — $\tfrac12\int(r_\text{out}^2-r_\text{in}^2)\,d\theta$。
Over the interval, the outer curve is the one with the... · 在该区间内，外曲线是具有...的曲线。
Outer = farther from the origin = larger $r$. · 外曲线 = 离原点更远 = 更大的$r$。
Square first, then subtract
- Just like Cartesian washers, it's $r_{\text{out}}^2-r_{\text{in}}^2$, not $(r_{\text{out}}-r_{\text{in}})^2$.
- Square each radius before subtracting.
- Check which curve is outer on the interval — it can swap, needing a split.
- The $\tfrac12$ and the squares are both essential.
先平方,再相减
- 和直角坐标垫圈一样,是 $r_{\text{out}}^2-r_{\text{in}}^2$,而非 $(r_{\text{out}}-r_{\text{in}})^2$。
- 相减前把每个半径平方。
- 检查区间上哪条曲线在外——它可能交换,需要分段。
- $\tfrac12$ 和平方都是必需的。
Area inside $r=2$, outside $r=1$: $\tfrac12\int_0^{2\pi}(4-1)\,d\theta$ (as a multiple of $\pi$). · 在$r=2$内、在$r=1$外的面积：$\tfrac12\int_0^{2\pi}(4-1)\,d\theta$ (以$\pi$的倍数形式填写)。
$\tfrac12\cdot3\cdot2\pi=3\pi$.
You square each radius before subtracting, not $(r_\text{out}-r_\text{in})^2$. · 应先分别对半径平方再相减，而非$(r_\text{out}-r_\text{in})^2$。
It is · 它是 $r_\text{out}^2-r_\text{in}^2$. · 它是 $r_\text{out}^2-r_\text{in}^2$。
If the outer and inner curves swap over the interval, you should split the integral at the crossing. · 如果在外部和内部曲线之间交换区间，你应该在交叉点分割积分。
A split keeps outer-minus-inner correct on each piece. · 分段确保每一部分都正确执行“外减内”。
The integrand is $r_{\text{outer}}^2-r_{\text{inner}}^2$ — square before subtracting, never $(r_{\text{out}}-r_{\text{in}})^2$. Get the intersection angles right (sketch the curves), and confirm which curve is outer over the interval; if they trade places, split the integral at the crossing.
被积函数是 $r_{\text{outer}}^2-r_{\text{inner}}^2$——相减前平方,绝非 $(r_{\text{out}}-r_{\text{in}})^2$。把交点角度求对(画曲线),并确认区间上哪条在外;若它们交换位置,就在交点处把积分分段。
Find the area inside $r=2$ but outside $r=1$ (an annulus).
- Outer $r=2$, inner $r=1$, over all angles $0$ to $2\pi$.
- $A=\dfrac12\displaystyle\int_0^{2\pi}\big(2^2-1^2\big)\,d\theta=\dfrac12\int_0^{2\pi}3\,d\theta=\dfrac12\cdot3\cdot2\pi=3\pi$.
- (Matches the ring area $\pi(2^2-1^2)=3\pi$.) ✓
求 $r=2$ 内但 $r=1$ 外的面积(一个圆环)。
- 外 $r=2$,内 $r=1$,遍及所有角度 $0$ 到 $2\pi$。
- $A=\dfrac12\displaystyle\int_0^{2\pi}\big(2^2-1^2\big)\,d\theta=\dfrac12\int_0^{2\pi}3\,d\theta=\dfrac12\cdot3\cdot2\pi=3\pi$。
- (与圆环面积 $\pi(2^2-1^2)=3\pi$ 一致。)✓
The area between two polar curves is $A=\frac{1}{2}\int_{\alpha}^{\beta}\big(r_{\text{outer}}^2-r_{\text{inner}}^2\big)\,d\theta$. First find the intersection angles for the limits, identify outer vs. inner, and square each radius before subtracting — never $(r_{\text{out}}-r_{\text{in}})^2$.
两条极曲线之间的面积是 $A=\frac{1}{2}\int_{\alpha}^{\beta}\big(r_{\text{outer}}^2-r_{\text{inner}}^2\big)\,d\theta$。先找交点角度作边界,判断外与内,并相减前把每个半径平方——绝非 $(r_{\text{out}}-r_{\text{in}})^2$。