Washer Method
A disk with a hole punched through it is called a washer. When you rotate the region between two curves, the inner curve carves out the hole. The washer volume formula is simply outer disk minus inner disk: $V = \pi\displaystyle\int_a^b \!\big([f(x)]^2 - [g(x)]^2\big)\,dx$. This lesson builds on the disk method and adds the subtraction step.
The region between $y = 4$ and $y = x^2$ (from $x = 0$ to $x = 2$) is rotated about the $x$-axis. Without a formula — explain in your own words why the volume of the resulting solid is not just $\pi\displaystyle\int_0^2 (4 - x^2)^2\,dx$.
When you rotate a single curve about an axis, each cross-section is a disk — a solid circle. When you rotate the region between two curves, the inner curve carves out a hole, making each cross-section a washer (an annulus).
At position $x$, the outer curve $y = f(x)$ has radius $R = f(x)$ and the inner curve $y = g(x)$ has radius $r = g(x)$. The washer area is:
$A(x) = \pi R^2 - \pi r^2 = \pi\big([f(x)]^2 - [g(x)]^2\big)$
Summing these washers over $[a,b]$:
$V = \pi\displaystyle\int_a^b \big([f(x)]^2 - [g(x)]^2\big)\,dx$
Critical point: subtract the squares, not the square of the difference.
$[f(x)]^2 - [g(x)]^2 \neq [f(x) - g(x)]^2$ in general.
Key facts
- Washer formula: $V = \pi\displaystyle\int_a^b\big([f(x)]^2 - [g(x)]^2\big)\,dx$
- $f(x)$ is the outer curve (larger radius), $g(x)$ is the inner curve (smaller radius)
- The limits $a$ and $b$ are found by solving $f(x) = g(x)$
Concepts
- Why the formula subtracts squares rather than squaring the difference
- How the inner curve carves a cylindrical hole from the solid
- The connection between the washer method and the disk method (disk is a special washer with $g = 0$)
Skills
- Identify outer and inner curves from a sketch and from the equations
- Find the limits of integration by solving $f(x) = g(x)$
- Evaluate the washer integral for polynomial, radical, and linear curves
Suppose two curves $y = f(x)$ (outer) and $y = g(x)$ (inner) bound a region, and this region is rotated about the $x$-axis. At position $x$, the cross-section is a washer with:
- Outer radius $R = f(x)$ → outer disk area $= \pi[f(x)]^2$
- Inner radius $r = g(x)$ → inner disk area $= \pi[g(x)]^2$
- Washer area $A(x) = \pi[f(x)]^2 - \pi[g(x)]^2$
Integrating from $a$ to $b$ gives the washer formula:
Key warning:
$[f(x)]^2 - [g(x)]^2 \neq [f(x) - g(x)]^2$
For example: $16 - 4 = 12$ but $(4-2)^2 = 4$. The difference-squared form is always wrong for the washer method.
Standard procedure:
- Sketch the two curves and shade the region being rotated.
- Identify the outer function $f(x)$ (farther from axis) and inner function $g(x)$.
- Find the limits: solve $f(x) = g(x)$ for the intersection $x$-values.
- Set up: $V = \pi\displaystyle\int_a^b\big([f(x)]^2 - [g(x)]^2\big)\,dx$.
- Expand, integrate term by term, and evaluate.
Suppose two curves $y = f(x)$ (outer) and $y = g(x)$ (inner) bound a region, and this region is rotated about the $x$-axis. At position $x$, the cross-section is a washer with:
Pause — copy the washer formula $V=\pi\int_a^b\{[f(x)]^2-[g(x)]^2\}\,dx$ and explain in one sentence why it equals outer disk minus inner disk into your book.
Quick check: The region between $y = 3$ (outer) and $y = x$ (inner) from $x = 0$ to $x = 3$ is rotated about the $x$-axis. Which integral gives the volume?
We just saw that the washer method gives $V=\pi\int_a^b\{[f(x)]^2-[g(x)]^2\}\,dx$ by subtracting the inner disk from the outer disk at each $x$. That raises a question: how do you reliably identify which curve is outer vs inner, and how do intersection points give you the correct limits? This card answers it → the outer curve has greater $|y|$ at any test $x$; limits come from solving $f(x)=g(x)$.
The two most important setup steps are (1) correctly identifying which curve is outer and (2) finding the integration limits.
Step 1 — Which curve is outer?
For rotation about the x-axis, the outer curve has the larger $y$-value (farther from the $x$-axis). Check at a specific $x$-value: if $f(x) > g(x) > 0$ for all $x \in [a,b]$, then $f$ is outer.
Step 2 — Finding limits:
Set $f(x) = g(x)$ and solve. These intersection $x$-values are typically $a$ and $b$.
Example: Region between $y = \sqrt{x}$ and $y = x^2$ rotated about the $x$-axis.
- Intersections: $\sqrt{x} = x^2 \Rightarrow x = x^4 \Rightarrow x^4 - x = 0 \Rightarrow x(x^3-1) = 0 \Rightarrow x = 0$ or $x = 1$.
- On $[0,1]$: check $x = 0.5$: $\sqrt{0.5} \approx 0.71 > 0.25 = (0.5)^2$, so $y = \sqrt{x}$ is outer.
- Integral: $V = \pi\displaystyle\int_0^1\big([\sqrt{x}]^2 - [x^2]^2\big)\,dx = \pi\displaystyle\int_0^1(x - x^4)\,dx$.
The two most important setup steps are (1) correctly identifying which curve is outer and (2) finding the integration limits.
Pause — copy the two setup steps: (1) test a point to identify outer/inner; (2) solve $f(x)=g(x)$ for the integration limits into your book.
Did you get this? True or false: for the region between $y = \sqrt{x}$ and $y = x^2$ rotated about the $x$-axis, the washer integrand simplifies to $x - x^4$.
Worked examples · 3 in a row, reveal as you go
The region between $y = x$ and $y = x^2$ (for $x \geq 0$) is rotated about the $x$-axis. Find the exact volume.
Find the volume when the region between $y = \sqrt{x}$ (outer) and $y = x^2$ (inner) for $0 \leq x \leq 1$ is rotated about the $x$-axis.
The region between $y = 4$ and $y = x^2$ for $-2 \leq x \leq 2$ is rotated about the $x$-axis. Find the exact volume.
Fill the gap: For the region between $y = 3$ (outer) and $y = x$ (inner) rotated about the $x$-axis, the integrand $[f(x)]^2 - [g(x)]^2$ equals .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the expression $[f(x)-g(x)]^2$ is equal to $[f(x)]^2 - [g(x)]^2$ when $f(x)$ and $g(x)$ are any two functions.
Activities · practice with the ideas
Find the volume when the region between $y = x$ and $y = x^2$ (for $x \geq 0$) is rotated about the $x$-axis. State the limits and show the integrand.
The region enclosed by $y = 2\sqrt{x}$ (outer) and $y = x$ (inner) for $0 \leq x \leq 4$ is rotated about the $x$-axis. Find the exact volume.
Find the volume when the region between $y = 2$ and $y = x^2 - 2$ (for $-2 \leq x \leq 2$) is rotated about the $x$-axis. (Check: which curve is outer?)
Show algebraically why $[f(x)]^2 - [g(x)]^2 \neq [f(x) - g(x)]^2$ by expanding both sides for $f(x) = x$ and $g(x) = x^2$.
The curves $y = x^2$ and $y = 4x - x^2$ intersect at $x = 0$ and $x = 2$. The region between them is rotated about the $x$-axis. Set up (but do not evaluate) the volume integral, identifying the outer curve.
Odd one out: For the region between $y = 2$ (outer) and $y = x$ (inner) rotated about the $x$-axis on $[0,2]$, three of these statements are correct. Which one is NOT?
At the start you predicted whether the volume formula for the region between two curves uses "outer disk minus inner disk". The answer is yes — but specifically it's $\pi[f(x)]^2 - \pi[g(x)]^2$ (subtract areas of the separate disks), not $\pi[f(x)-g(x)]^2$ (square the difference of the radii).
The key geometric insight: the hole in each washer has area $\pi r^2 = \pi[g(x)]^2$. You remove that hole from the full disk area $\pi R^2 = \pi[f(x)]^2$. Subtracting the cross-terms (which appear in $(f-g)^2$) would count or ignore wrong regions.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. The region between $y = x$ (outer) and $y = x^2$ (inner) for $0 \leq x \leq 1$ is rotated about the $x$-axis. Find the exact volume. (2 marks)
Q2. Find the volume of the solid formed when the region enclosed by $y = \sqrt{x}$ and $y = x^2$ (for $0 \leq x \leq 1$) is rotated about the $x$-axis. (3 marks)
Q3. Explain why $V = \pi\displaystyle\int_a^b [f(x)-g(x)]^2\,dx$ is the wrong formula for the washer method. Include an algebraic demonstration using $f(x) = 3$ and $g(x) = x$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $V = \dfrac{2\pi}{15}$ (shown in worked example 1) · 2. $[2\sqrt{x}]^2 = 4x$, $[x]^2 = x^2$; $V = \pi\displaystyle\int_0^4(4x-x^2)\,dx = \pi\Big[2x^2-\dfrac{x^3}{3}\Big]_0^4 = \pi\Big(32-\dfrac{64}{3}\Big) = \dfrac{32\pi}{3}$ · 3. $y=2$ is outer on $[-2,2]$; $[2]^2=4$, $[x^2-2]^2 = x^4-4x^2+4$; integrand $= 4-(x^4-4x^2+4) = 4x^2-x^4$; by symmetry $V = 2\pi\displaystyle\int_0^2(4x^2-x^4)\,dx = 2\pi\Big[\dfrac{4x^3}{3}-\dfrac{x^5}{5}\Big]_0^2 = 2\pi\Big(\dfrac{32}{3}-\dfrac{32}{5}\Big) = 2\pi\cdot\dfrac{64}{15} = \dfrac{128\pi}{15}$ · 4. LHS $= x^2-x^4$; RHS $= (x-x^2)^2 = x^2-2x^3+x^4$. Different by $-2x^3$ and $+2x^4$ terms · 5. $V = \pi\displaystyle\int_0^2\big((4x-x^2)^2 - x^4\big)\,dx = \pi\displaystyle\int_0^2\big(16x^2-8x^3+x^4-x^4\big)\,dx = \pi\displaystyle\int_0^2(16x^2-8x^3)\,dx$.
Q1 (2 marks): Integrand $= x^2 - x^4$ [1]; $V = \pi\Big[\dfrac{x^3}{3}-\dfrac{x^5}{5}\Big]_0^1 = \pi\Big(\dfrac{1}{3}-\dfrac{1}{5}\Big) = \dfrac{2\pi}{15}$ units$^3$ [1].
Q2 (3 marks): Outer: $y=\sqrt{x}$, inner: $y=x^2$ (verified at $x=0.5$) [1]; $[\sqrt{x}]^2-[x^2]^2 = x-x^4$ [1]; $V = \pi\Big[\dfrac{x^2}{2}-\dfrac{x^5}{5}\Big]_0^1 = \pi\Big(\dfrac{1}{2}-\dfrac{1}{5}\Big) = \dfrac{3\pi}{10}$ units$^3$ [1].
Q3 (3 marks): The washer cross-section has area $\pi R^2 - \pi r^2$, not $\pi(R-r)^2$ — the hole punched out has area $\pi r^2$, not $\pi(R-r)^2$ [1]. With $f=3$, $g=x$: correct integrand $= 9-x^2$; wrong formula gives $(3-x)^2 = 9-6x+x^2$ [1]. These differ by $-6x+2x^2$, which has no geometric interpretation — it includes a cross-term $-6x$ that does not correspond to any physical area being added or removed [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering washer method questions. Lighter alternative to the boss.
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