Volume with Washer Method: Revolving Around Other Axes
Введено| English | Значение |
|---|---|
| outer/ˈaʊtə/ | внешний |
| inner/ˈɪnə/ | внутренний |
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.
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.
Washers about y = −1: outer and inner radii are both measured from the axis
Две кривые над смещенной осью
y = ax² + bx
Revolving about $y=-1$, both radii grow by $1$ — outer $\sqrt x+1$, inner $x+1$; still $R^2-r^2$.
Revolving the region between $y=\sqrt x$ и $y=x$ about $y=-1$, the outer and inner radii are...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Shift оба варианта radii by $+1$ down to $y=-1$.
For a shifted axis, you must re-measure ____ radii (outer and inner) to the new line.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Shift both, not just one.
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.
After shifting the axis, you should...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The axis position can swap outer/inner roles.
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.
Even about a shifted axis, the washer integrand is $R^2-r^2$, not $(R-r)^2$.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Always square each radius before subtracting.
The volume for $R=\sqrt x+1$, $r=x+1$ on $[0,1]$ about $y=-1$ is...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
$\pi\int_0^1(2\sqrt x-x-x^2)\,dx=\pi(\tfrac43-\tfrac12-\tfrac13)=\tfrac{\pi}{2}$.
The shifted-washer method combines which two ideas?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Washer (gap → ring) plus a shifted radius.
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.
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 $r=x+1$.
- $V=\pi\displaystyle\int_0^1\Big((\sqrt x+1)^2-(x+1)^2\Big)\,dx=\pi\int_0^1\big(x+2\sqrt x - x^2 - 2x\big)\,dx$.
- $=\pi\int_0^1\big(2\sqrt x - x - x^2\big)\,dx=\pi\big(\tfrac43-\tfrac12-\tfrac13\big)=\tfrac{\pi}{2}$.
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 $(R-r)^2$.
Write from washers about a shifted axis
- A region lies above y = 0 and is revolved about y = -2.
- Both the outer and inner edges move farther from that axis.
How do the two radii change, and what must you not do?
Both edges are measured to the same shifted line.
Washers about a line 2 units below the region change the radii how?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The shift lengthens every radius to that line.