Volume with Washer Method: Revolving Around the x- or y-Axis
Введено| English | Значение |
|---|---|
| washer/ˈwɒʃə/ | кольцо (шайба) |
| washer method/ˈwɒʃə ˈmeθəd/ | метод колец (шайб) |
| inner radius/ˈɪnə ˈreɪdɪəs/ | внутренний радиус |
| outer radius/ˈaʊtə ˈreɪdɪəs/ | внешний радиус |
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.
A washer (rather than a disc) appears when...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The gap leaves a hole → a washer.
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.
Две кривые образуют кольцо
y = ax² + bx
Between $y=x$ и $y=\sqrt x$ there is a gap from the axis — revolving makes washers with outer $\sqrt x$, inner $x$.
The washer-method volume is...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Big circle minus small circle: $\pi(R^2-r^2)$.
In a washer, $R$ is the ____ radius (to the far curve) and $r$ the inner radius.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Outer to the far curve, inner to the near curve.
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.
Revolving the region between $y=\sqrt x$ (top) and $y=x$ (bottom) about the $x$-axis, the outer radius is...
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
The farther curve from the axis, $\sqrt x$, is the outer radius.
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.
The washer integrand $R^2-r^2$ is the same as $(R-r)^2$.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Square each radius first; $(R-r)^2$ is wrong.
За $R=\sqrt x$, $r=x$ on $[0,1]$: $V=\pi\int_0^1 (x-x^2)\,dx=$
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
$\pi(\tfrac12-\tfrac13)=\tfrac{\pi}{6}$.
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.
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 $[0,1]$, $\sqrt x\ge x$).
- $V=\pi\displaystyle\int_0^1\big((\sqrt x)^2-x^2\big)\,dx=\pi\int_0^1 (x-x^2)\,dx=\pi\Big(\tfrac12-\tfrac13\Big)=\tfrac{\pi}{6}$.
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 $(R-r)^2$.
Write from a washer volume of 9 pi
- The outer radius is 2 and the inner radius is 1.
- The washers run from x = 0 to x = 3.
What is the volume?
pi times 3 times 3 is 9 pi.
Outer radius 2, inner radius 1, from 0 to 3. The volume is what?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
pi (4 - 1) times 3 = 9 pi.