Volume with Washer Method: Revolving Around Other Axes · 垫圈法体积：绕其他轴旋转
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| outer/ˈaʊtə/ | 外 | wài |
| inner/ˈɪnə/ | 内 | nèi |
Washers around a shifted line
- The last piece: revolve a two-curve region around a line that isn't the $x$- or $y$-axis.
- Combine the two ideas you just learned: washers (a gap → a ring) and a shifted axis (adjust the radius).
- The formula stays $V=\pi\int\big(R^2-r^2\big)$; only $R$ and $r$ get re-measured to the new line.
- Get both distances right and the calculus is routine.
绕平移线的垫圈
- 最后一块:把双曲线区域绕一条不是 $x$ 轴或 $y$ 轴的线旋转。
- 结合你刚学的两个想法:垫圈(有间隙 → 环)和平移轴(调整半径)。
- 公式仍是 $V=\pi\int\big(R^2-r^2\big)$;只有 $R$ 和 $r$ 要重新量到新线。
- 把两个距离都求对,微积分就是例行公事。
Both radii measured to the new axis
- For a horizontal axis $y=k$: $R=|\text{far curve}-k|$ and $r=|\text{near curve}-k|$.
- Each radius is the distance from its curve to the shifted line, not to $y=0$.
- The outer 外 radius reaches the curve farther from the axis; the inner 内 radius the nearer one.
- Then $V=\pi\int\big(R^2-r^2\big)$ as usual.
两个半径都量到新轴
- 对水平轴 $y=k$:$R=|\text{far curve}-k|$,$r=|\text{near curve}-k|$。
- 每个半径是从它的曲线到平移线的距离,而非到 $y=0$。
- 外半径到离轴更远的曲线;内半径到更近的那条。
- 然后照常 $V=\pi\int\big(R^2-r^2\big)$。
Two curves above a shifted axis · 平移轴上方的两条曲线
y = ax² + bx
Revolving about $y=-1$, both radii grow by $1$ — outer $\sqrt x+1$, inner $x+1$; still $R^2-r^2$. · 绕$y=-1$旋转，两个半径都增加$1$——外$\sqrt x+1$，内$x+1$；仍然是$R^2-r^2$。
Revolving the region between $y=\sqrt x$ and $y=x$ about $y=-1$, the outer and inner radii are... · 将 $y=\sqrt x$ 和 $y=x$ 之间的区域绕 $y=-1$ 旋转，外半径和内半径分别为...
Shift · 平移 both · 两者 radii by $+1$ down to $y=-1$. · 将两个半径都向下平移$+1$至$y=-1$。
For a shifted axis, you must re-measure ____ radii (outer and inner) to the new line. · 对于平移后的轴，必须重新测量到直线的____半径（外半径和内半径）。
Shift both, not just one. · 平移两个，而不只是一个。
Watch which curve is now farther
- Shifting the axis can swap which curve is outer vs. inner — re-check after moving the line.
- Revolving about a line below the region: the lower curve is now the inner one (nearer the axis).
- Revolving about a line above the region: the roles flip.
- Always identify outer/inner relative to the actual axis of revolution.
注意现在哪条曲线更远
- 平移轴可能交换哪条曲线是外、哪条是内——移线后重新检查。
- 绕区域下方的线旋转:下方曲线现在是内侧(更近轴)。
- 绕区域上方的线旋转:角色反转。
- 永远相对于实际旋转轴来识别外/内。
After shifting the axis, you should... · 平移轴后，你应该...
The axis position can swap outer/inner roles. · 轴的位置可以交换内外半径的角色。
Still: square each radius first
- As with any washer, subtract $R^2-r^2$ — square the (shifted) radii before subtracting.
- Never $(R-r)^2$, and never reuse the plain-axis radii.
- Draw the two distance segments from the axis to each curve to keep them straight.
- Then integrate along the appropriate variable.
仍然:先把每个半径平方
- 与任何垫圈一样,相减 $R^2-r^2$——相减前把(平移后的)半径平方。
- 绝非 $(R-r)^2$,也绝不沿用坐标轴的半径。
- 画出从轴到每条曲线的两条距离线段以保持清晰。
- 然后沿合适的变量积分。
Even about a shifted axis, the washer integrand is $R^2-r^2$, not $(R-r)^2$. · 即使绕平移后的轴，垫圈被积函数仍是$R^2-r^2$，而不是$(R-r)^2$。
Always square each radius before subtracting. · 相减前务必先对每个半径平方。
The volume for $R=\sqrt x+1$, $r=x+1$ on $[0,1]$ about $y=-1$ is... · $R=\sqrt x+1$、$r=x+1$在$[0,1]$上绕$y=-1$旋转的体积为...
$\pi\int_0^1(2\sqrt x-x-x^2)\,dx=\pi(\tfrac43-\tfrac12-\tfrac13)=\tfrac{\pi}{2}$.
The shifted-washer method combines which two ideas? · 平移垫圈法结合了哪两个概念？
Washer (gap → ring) plus a shifted radius. · 垫圈（间隙→环）加上平移半径。
Two combined pitfalls here: (1) shift both radii to the new axis (not just one), and (2) still use $R^2-r^2$, not $(R-r)^2$. After shifting, re-check which curve is outer — the axis's position can swap the roles. Sketch both radius segments to the actual line before squaring.
这里有两个叠加的陷阱:(1) 把两个半径都平移到新轴(不只是一个),(2) 仍用 $R^2-r^2$,而非 $(R-r)^2$。平移后,重新检查哪条曲线是外侧——轴的位置可能交换角色。平方前画出到实际线的两条半径线段。
Revolve the region between $y=\sqrt x$ (top) and $y=x$ (bottom) on $[0,1]$ about the line $y=-1$.
- Shift both radii down to $y=-1$: outer $R=\sqrt x+1$, inner $r=x+1$.
- $V=\pi\displaystyle\int_0^1\Big((\sqrt x+1)^2-(x+1)^2\Big)\,dx=\pi\int_0^1\big(x+2\sqrt x - x^2 - 2x\big)\,dx$.
- $=\pi\int_0^1\big(2\sqrt x - x - x^2\big)\,dx=\pi\big(\tfrac43-\tfrac12-\tfrac13\big)=\tfrac{\pi}{2}$.
把 $[0,1]$ 上 $y=\sqrt x$(上)与 $y=x$(下)之间的区域绕线 $y=-1$ 旋转。
- 把两个半径都向下移到 $y=-1$:外 $R=\sqrt x+1$,内 $r=x+1$。
- $V=\pi\displaystyle\int_0^1\Big((\sqrt x+1)^2-(x+1)^2\Big)\,dx=\pi\int_0^1\big(x+2\sqrt x - x^2 - 2x\big)\,dx$。
- $=\pi\int_0^1\big(2\sqrt x - x - x^2\big)\,dx=\pi\big(\tfrac43-\tfrac12-\tfrac13\big)=\tfrac{\pi}{2}$。
The washer method about a shifted axis: $V=\pi\int\big(R^2-r^2\big)$, but re-measure both the outer and inner radii to that line (e.g. add the shift for $y=k$ below the region). Re-check which curve is outer, and always square each radius before subtracting — never $(R-r)^2$.
绕平移轴的垫圈法:$V=\pi\int\big(R^2-r^2\big)$,但把外半径和内半径都重新量到那条线(如对区域下方的 $y=k$ 加上平移量)。重新检查哪条曲线是外侧,并永远在相减前把每个半径平方——绝非 $(R-r)^2$。