Volume with Washer Method: Revolving Around the x- or y-Axis · 垫圈法体积：绕 x 轴或 y 轴旋转
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| washer/ˈwɒʃə/ | 垫圈 | diàn juàn |
| washer method/ˈwɒʃə ˈmeθəd/ | 垫圈法 | diàn juàn fǎ |
| inner radius/ˈɪnə ˈreɪdɪəs/ | 内半径 | nèi bàn jìng |
| outer radius/ˈaʊtə ˈreɪdɪəs/ | 外半径 | wài bàn jìng |
A disc with a hole in it
- When you revolve a region that has a gap between two curves, each slice isn't a solid disc — it's a ring.
- That ring (a disc with a circular hole punched out) is a washer 垫圈.
- Its area is the big circle minus the small circle: $\pi R^2-\pi r^2$.
- Adding up washers is the washer method 垫圈法.
中间有洞的圆盘
- 当你旋转一个在两曲线之间有间隙的区域时,每片不是实心圆盘——而是一个环。
- 那个环(挖掉一个圆洞的圆盘)是一个垫圈。
- 它的面积是大圆减小圆:$\pi R^2-\pi r^2$。
- 把垫圈加起来就是垫圈法。
A washer (rather than a disc) appears when... · 出现垫圈（而非圆盘）的情况是...
The gap leaves a hole → a washer. · 间隙留下孔洞 → 垫圈。
Outer minus inner radius
- Each washer has an outer radius 外半径 $R$ (to the far curve) and an inner radius 内半径 $r$ (to the near curve).
- Its area is $\pi\big(R^2-r^2\big)$, so:
-
$$V=\pi\int_a^b \Big(\big[R(x)\big]^2-\big[r(x)\big]^2\Big)\,dx$$
- Revolving about the $x$-axis, $R$ and $r$ are the distances from the two curves to the axis.
外半径减内半径
- 每个垫圈有一个外半径 $R$(到远曲线)和一个内半径 $r$(到近曲线)。
- 它的面积是 $\pi\big(R^2-r^2\big)$,所以:
-
$$V=\pi\int_a^b \Big(\big[R(x)\big]^2-\big[r(x)\big]^2\Big)\,dx$$
- 绕 $x$ 轴旋转,$R$ 和 $r$ 是两曲线到轴的距离。
Two curves make a ring · 两条曲线形成一个环
y = ax² + bx
Between $y=x$ and $y=\sqrt x$ there is a gap from the axis — revolving makes washers with outer $\sqrt x$, inner $x$. · 在$y=x$和$y=\sqrt x$之间有一个远离轴的间隙——旋转产生外径为$\sqrt x$、内径为$x$的垫圈。
The washer-method volume is... · 垫圈法的体积为...
Big circle minus small circle: $\pi(R^2-r^2)$. · 大圆减小圆：$\pi(R^2-r^2)$。
In a washer, $R$ is the ____ radius (to the far curve) and $r$ the inner radius. · 在圆环体中，$R$ 是 ____ 半径（到远曲线），$r$ 是内半径。
Outer to the far curve, inner to the near curve. · 外半径对应远曲线，内半径对应近曲线。
Identifying $R$ and $r$
- $R$ (outer) comes from the curve farther from the axis; $r$ (inner) from the nearer curve.
- Revolving the region between $y=f(x)$ (top) and $y=g(x)$ (bottom) about the $x$-axis: $R=f(x)$, $r=g(x)$.
- The hole exists because the bottom curve doesn't reach the axis.
- Square each radius separately before subtracting.
识别 $R$ 与 $r$
- $R$(外)来自更远离轴的曲线;$r$(内)来自更近的曲线。
- 把 $y=f(x)$(上)与 $y=g(x)$(下)之间的区域绕 $x$ 轴旋转:$R=f(x)$,$r=g(x)$。
- 洞的存在是因为下方曲线不到达轴。
- 相减前把每个半径分别平方。
Revolving the region between $y=\sqrt x$ (top) and $y=x$ (bottom) about the $x$-axis, the outer radius is... · 将$y=\sqrt x$（上）和$y=x$（下）之间的区域绕$x$轴旋转，外半径为...
The farther curve from the axis, $\sqrt x$, is the outer radius. · 离轴较远的曲线$\sqrt x$是外半径。
Square first, then subtract
- The area is $\pi(R^2-r^2)$, not $\pi(R-r)^2$ — square each radius before you subtract.
- $(R-r)^2$ would wrongly include a cross term and give the wrong volume.
- So compute $R^2$ and $r^2$ each, subtract, then integrate and multiply by $\pi$.
- This is the single most common washer mistake.
先平方,再相减
- 面积是 $\pi(R^2-r^2)$,而非 $\pi(R-r)^2$——相减前先把每个半径平方。
- $(R-r)^2$ 会错误地引入一个交叉项,给出错误体积。
- 所以各算 $R^2$ 和 $r^2$,相减,再积分并乘以 $\pi$。
- 这是垫圈法最常见的单一错误。
The washer integrand $R^2-r^2$ is the same as $(R-r)^2$. · 垫圈被积函数$R^2-r^2$与$(R-r)^2$相同。
Square each radius first; $(R-r)^2$ is wrong. · 先对每个半径平方；$(R-r)^2$是错误的。
For · 支持 $R=\sqrt x$, $r=x$ on $[0,1]$: $V=\pi\int_0^1 (x-x^2)\,dx=$ · 对于$R=\sqrt x$，$r=x$在$[0,1]$上：$V=\pi\int_0^1 (x-x^2)\,dx=$
$\pi(\tfrac12-\tfrac13)=\tfrac{\pi}{6}$.
The washer integrand is $R^2-r^2$, not $(R-r)^2$ — square the radii before subtracting. And keep $R$ (outer, to the far curve) distinct from $r$ (inner, to the near curve); swapping them, or using $(R-r)^2$, both give wrong volumes. A washer appears only when a gap separates the region from the axis.
垫圈被积函数是 $R^2-r^2$,而非 $(R-r)^2$——相减前把半径平方。并保持 $R$(外,到远曲线)与 $r$(内,到近曲线)不同;交换它们或用 $(R-r)^2$ 都给出错误体积。只有当间隙把区域与轴分开时才出现垫圈。
Revolve the region between $y=x$ (bottom) and $y=\sqrt x$ (top) on $[0,1]$ about the $x$-axis.
- Outer radius $R=\sqrt x$, inner radius $r=x$ (on $[0,1]$, $\sqrt x\ge x$).
- $V=\pi\displaystyle\int_0^1\big((\sqrt x)^2-x^2\big)\,dx=\pi\int_0^1 (x-x^2)\,dx=\pi\Big(\tfrac12-\tfrac13\Big)=\tfrac{\pi}{6}$.
把 $[0,1]$ 上 $y=x$(下)与 $y=\sqrt x$(上)之间的区域绕 $x$ 轴旋转。
- 外半径 $R=\sqrt x$,内半径 $r=x$(在 $[0,1]$ 上,$\sqrt x\ge x$)。
- $V=\pi\displaystyle\int_0^1\big((\sqrt x)^2-x^2\big)\,dx=\pi\int_0^1 (x-x^2)\,dx=\pi\Big(\tfrac12-\tfrac13\Big)=\tfrac{\pi}{6}$。
The washer method revolves a region with a gap into rings: $V=\pi\int_a^b\big(R^2-r^2\big)\,dx$, where $R$ is the outer radius (far curve) and $r$ the inner radius (near curve). Square each radius before subtracting — $R^2-r^2$, never $(R-r)^2$.
垫圈法把有间隙的区域旋转成环:$V=\pi\int_a^b\big(R^2-r^2\big)\,dx$,其中 $R$ 是外半径(远曲线),$r$ 是内半径(近曲线)。相减前把每个半径平方——是 $R^2-r^2$,绝非 $(R-r)^2$。