Volume with Disc Method: Revolving Around Other Axes · 圆盘法体积：绕其他轴旋转
Spinning around a shifted line
- What if you revolve a region not around an axis, but around a line like $y=-1$ or $x=3$?
- The disc method still works — you just measure the radius from the curve to that shifted axis.
- The formula $V=\pi\int R^2$ is unchanged; only $R$ needs adjusting.
- Getting the radius right is the whole game.
绕一条平移的线旋转
- 如果不是绕坐标轴,而是绕像 $y=-1$ 或 $x=3$ 这样的线旋转呢?
- 圆盘法仍然有效——你只需从曲线到那条平移轴测量半径。
- 公式 $V=\pi\int R^2$ 不变;只有 $R$ 需要调整。
- 把半径求对就是全部关键。
Radius = distance to the new axis
- Revolving about a horizontal line $y=k$: the radius is the vertical gap $R=|f(x)-k|$.
- Revolving about a vertical line $x=k$: the radius is the horizontal gap $R=|g(y)-k|$.
- Subtract the axis's value from the curve's value (in the right variable) to get the distance.
- Then square, multiply by $\pi$, and integrate as usual.
半径 = 到新轴的距离
- 绕水平线 $y=k$ 旋转:半径是竖直间隙 $R=|f(x)-k|$。
- 绕竖直线 $x=k$ 旋转:半径是水平间隙 $R=|g(y)-k|$。
- 用曲线的值减去轴的值(在正确的变量里)得到距离。
- 然后平方,乘以 $\pi$,像往常一样积分。
The curve above a shifted axis · 平移轴上方的曲线
y = a·√x
Revolving about $y=-1$ makes the radius reach from the curve down to that line — $R=f(x)+1$, larger than $f(x)$. · 绕$y=-1$旋转会使半径从曲线延伸到该直线——$R=f(x)+1$，大于$f(x)$。
Revolving $y=f(x)$ about the line $y=-1$, the radius is... · 绕 $y=f(x)$ 直线旋转，半径为… $y=-1$，半径是...
Distance from $f(x)$ down to $y=-1$ is $f(x)-(-1)=f(x)+1$. · 从$f(x)$向下到$y=-1$的距离是$f(x)-(-1)=f(x)+1$。
Revolving about $y=k$ (below the region), the radius of a curve $f(x)$ is... · 绕$y=k$（位于区域下方）旋转，曲线$f(x)$的半径为...
Curve value minus axis value: $f(x)-k$. · 曲线值减去轴值：$f(x)-k$。
Revolving about a vertical line $x=k$, the radius is the ____ distance $|g(y)-k|$. · 绕垂直线$x=k$旋转，半径是$|g(y)-k|$的____距离。
For a vertical axis, measure horizontally. · 对于垂直轴，测量水平距离。
Axis below the region
- If the axis $y=k$ is below the region, the curve is above it, so $R=f(x)-k$ (a positive gap).
- Example: revolving about $y=-1$ a curve at height $f(x)$ gives radius $R=f(x)-(-1)=f(x)+1$.
- The "$+1$" is the extra distance down to the shifted axis.
- Always draw it: the radius reaches from the curve to the axis.
轴在区域下方
- 若轴 $y=k$ 在区域下方,曲线在它上方,所以 $R=f(x)-k$(一个正间隙)。
- 例:把高度为 $f(x)$ 的曲线绕 $y=-1$ 旋转,半径 $R=f(x)-(-1)=f(x)+1$。
- 那个"$+1$"是向下到平移轴的额外距离。
- 永远画出来:半径从曲线延伸到轴。
If the axis $y=3$ lies above the region (curve $f(x)<3$), the radius is... · 如果轴$y=3$位于上方（曲线$f(x)<3$），半径为...
Distance is axis minus curve: $3-f(x)$ (a positive gap). · 距离是轴减去曲线：$3-f(x)$（正间隙）。
Don't reuse the plain-axis radius
- The most common error is forgetting to shift: using $R=f(x)$ when the axis is $y=-1$.
- That underestimates the radius by the shift amount, giving a wrong (too small) volume.
- Rewrite the radius as (curve value) $-$ (axis value) every time the axis isn't $y=0$ or $x=0$.
- Sketching the distance segment prevents this slip.
别沿用坐标轴的半径
- 最常见的错误是忘了平移:轴是 $y=-1$ 时却用 $R=f(x)$。
- 那把半径低估了平移量,给出错误(偏小)的体积。
- 每当轴不是 $y=0$ 或 $x=0$,就把半径改写成(曲线值)$-$(轴值)。
- 画出距离线段能防止这个失误。
When revolving about $y=-1$, you can still use radius $R=f(x)$. · 当绕$y=-1$旋转时，仍可使用半径$R=f(x)$。
You must shift: $R=f(x)+1$. · 你必须平移：$R=f(x)+1$。
Revolving $y=\sqrt x$ on $[0,4]$ about $y=-1$ gives radius $\sqrt x+1$, so $V=$ · 旋转 $y=\sqrt x$ 在 $[0,4]$ 关于 $y=-1$ 给出半径 $\sqrt x+1$, 所以 $V=$
$\pi\int_0^4(x+2\sqrt x+1)\,dx=\pi(8+\tfrac{32}{3}+4)=\tfrac{68\pi}{3}$.
For a shifted axis, the radius is the distance from the curve to that line, e.g. $R=f(x)-k$ for $y=k$ — not just $f(x)$. Revolving about $y=-1$ makes the radius larger (add $1$); about $y=3$ above the region makes it $3-f(x)$. Always measure to the correct axis before squaring.
对平移轴,半径是从曲线到那条线的距离,如对 $y=k$ 是 $R=f(x)-k$——而非只是 $f(x)$。绕 $y=-1$ 旋转使半径更大(加 $1$);绕区域上方的 $y=3$ 则是 $3-f(x)$。平方前永远量到正确的轴。
Revolve the region between $y=\sqrt x$ and the line $y=-1$ on $[0,4]$ about $y=-1$. (The region reaches down to the axis, so each slice sweeps a solid disc — no hole.)
- Radius reaches from the curve down to $y=-1$: $R(x)=\sqrt x-(-1)=\sqrt x+1$.
- $V=\pi\displaystyle\int_0^4 (\sqrt x+1)^2\,dx=\pi\int_0^4 \big(x+2\sqrt x+1\big)\,dx$.
- $=\pi\Big[\tfrac{x^2}{2}+\tfrac{4}{3}x^{3/2}+x\Big]_0^4=\pi\big(8+\tfrac{32}{3}+4\big)=\tfrac{68\pi}{3}$.
把 $[0,4]$ 上 $y=\sqrt x$ 与线 $y=-1$ 之间的区域绕 $y=-1$ 旋转。(区域一直向下到旋转轴,所以每一薄片扫出一个实心圆盘——没有孔。)
- 半径从曲线向下延伸到 $y=-1$:$R(x)=\sqrt x-(-1)=\sqrt x+1$。
- $V=\pi\displaystyle\int_0^4 (\sqrt x+1)^2\,dx=\pi\int_0^4 \big(x+2\sqrt x+1\big)\,dx$。
- $=\pi\Big[\tfrac{x^2}{2}+\tfrac{4}{3}x^{3/2}+x\Big]_0^4=\pi\big(8+\tfrac{32}{3}+4\big)=\tfrac{68\pi}{3}$。
The disc method about a shifted axis keeps $V=\pi\int R^2$, but the radius is the distance from the curve to that line: $R=f(x)-k$ for $y=k$, or $g(y)-k$ for $x=k$. Adjust $R$ for the shift (an axis below the region adds to the radius); never reuse the plain-axis radius.
绕平移轴的圆盘法保持 $V=\pi\int R^2$,但半径是从曲线到那条线的距离:对 $y=k$ 是 $R=f(x)-k$,对 $x=k$ 是 $g(y)-k$。为平移调整 $R$(区域下方的轴会增加半径);永远别沿用坐标轴的半径。