Volumes Between Curves — Worked Examples
The washer method extends naturally to regions enclosed by two curves. By subtracting the inner solid from the outer solid, $V = \pi\displaystyle\int_a^b\!\bigl[f(x)^2 - g(x)^2\bigr]\,dx$, you unlock a powerful technique for any solid with a hole through its axis. In this lesson you'll consolidate the method through multi-step worked examples and tackle increasingly complex regions.
The region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ is rotated about the $x$-axis. Without calculating — which curve gives the outer radius and which gives the inner radius? Explain your reasoning.
When a region between two curves is rotated about the $x$-axis, each slice is a washer — a disk with a hole. The volume is the outer disk minus the inner disk:
For $f(x) \geq g(x) \geq 0$ on $[a,b]$, rotating about the $x$-axis gives:
$V = \pi\!\displaystyle\int_a^b\!\bigl[f(x)^2 - g(x)^2\bigr]\,dx$
where $R = f(x)$ is the outer radius and $r = g(x)$ is the inner radius.
Key facts
- $V = \pi\displaystyle\int_a^b [f(x)^2 - g(x)^2]\,dx$ is the washer-method formula
- Outer radius $R = f(x)$ is the curve farther from the axis
- Limits of integration come from the intersection points of the two curves
Concepts
- Why each thin washer has area $\pi(R^2 - r^2)$ and how summing washers gives volume
- How to decide which curve is the outer radius for any given region
- The connection between the washer method and the disk method (washer = disk with hole)
Skills
- Set up and evaluate washer-method integrals for regions between two curves
- Find intersection points and use them as limits of integration
- Handle regions where the curves cross, splitting the integral appropriately
To find the volume of the solid formed when the region between $y = f(x)$ and $y = g(x)$ (with $f(x) \geq g(x) \geq 0$) is rotated about the $x$-axis, follow these four steps:
- Sketch both curves and identify the enclosed region.
- Find intersections: solve $f(x) = g(x)$ to get the limits $a$ and $b$.
- Identify radii: outer $R = f(x)$, inner $r = g(x)$.
- Integrate: $V = \pi\displaystyle\int_a^b\!\bigl[f(x)^2 - g(x)^2\bigr]\,dx$.
Example — the hook question: Region between $y = x$ and $y = x^2$, $0 \leq x \leq 1$, rotated about the $x$-axis.
On $[0,1]$: $x \geq x^2$ so $f(x) = x$ (outer), $g(x) = x^2$ (inner).
$V = \pi\displaystyle\int_0^1\!\bigl[x^2 - x^4\bigr]\,dx = \pi\left[\dfrac{x^3}{3} - \dfrac{x^5}{5}\right]_0^1 = \pi\!\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{2\pi}{15}$
To find the volume of the solid formed when the region between $y = f(x)$ and $y = g(x)$ (with $f(x) \geq g(x) \geq 0$) is rotated about the $x$-axis, follow these four steps:
Pause — copy the full four-step washer setup: (1) outer/inner identification; (2) find limits; (3) write $V=\pi\int_a^b([R]^2-[r]^2)\,dx$; (4) expand and integrate into your book.
Quick check: The region between $y = \sqrt{x}$ and $y = x$ (for $0 \leq x \leq 1$) is rotated about the $x$-axis. What is the outer radius?
We just saw the full washer setup: identify outer/inner curves, find limits, write $V=\pi\int_a^b([R(x)]^2-[r(x)]^2)\,dx$, evaluate. That raises a question: which standard curve pairs appear most often in HSC problems, and how do you pre-classify them so you can start the setup instantly? This card answers it → parabola-line, two parabolas, and line-parabola pairs each have predictable shapes; a quick sketch decides outer vs inner.
Many HSC problems involve standard curve pairs. Recognising the shape of each region speeds up setup considerably:
- Line and parabola: e.g. $y = x$ and $y = x^2$. Intersect at $x = 0$ and $x = 1$. Line is outer on $[0,1]$.
- Two parabolas: e.g. $y = \sqrt{x}$ and $y = x^2$. Intersect at $x = 0$ and $x = 1$. $\sqrt{x} \geq x^2$ on $[0,1]$.
- Trig and polynomial: e.g. $y = \cos x$ and $y = \sin x$. Find intersections, determine which is larger on each sub-interval.
Key step: always expand $(R^2 - r^2)$ before integrating — never try to integrate $(f - g)^2$ as a shortcut (that gives a different result).
Many HSC problems involve standard curve pairs. Recognising the shape of each region speeds up setup considerably:
Pause — copy the three common HSC curve pairs (parabola-line, two parabolas, line-parabola) and a sketch-level diagram showing which is outer for each 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 correct setup is $V = \pi\!\displaystyle\int_0^1\![x - x^4]\,dx$.
Worked examples · 3 in a row, reveal as you go
Find the volume of the solid formed when the region enclosed by $y = x$ and $y = x^2$ is rotated about the $x$-axis.
Find the volume when the region enclosed by $y = \sqrt{x}$ and $y = x^2$ is rotated about the $x$-axis.
The region enclosed by $x = y^2$ and $x = y$ (for $y \geq 0$) is rotated about the $y$-axis. Find the volume.
Fill the gap: For the region between $y = \sqrt{x}$ and $y = x^2$ rotated about the $x$-axis, $V = \pi\!\displaystyle\int_0^1\!\bigl[x -$ $\bigr]\,dx$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if the washer-method integrand $(R^2 - r^2)$ is negative over the integration interval, the outer and inner radii have been assigned incorrectly.
Activities · practice with the ideas
Find the intersections of $y = x$ and $y = x^3$, then state which curve is the outer radius on the positive interval.
Set up (do not evaluate) the washer-method integral for rotating the region between $y = \sqrt{x}$ and $y = x$ about the $x$-axis.
Evaluate $\pi\!\displaystyle\int_0^1\!\bigl[x - x^4\bigr]\,dx$ and state the volume in exact form.
Explain why the formula uses $(R^2 - r^2)$ rather than $(R - r)^2$. Use the example $R = 3$, $r = 1$ to illustrate.
The region between $y = 2x$ and $y = x^2$ is rotated about the $x$-axis. Find the volume.
Odd one out: Three of these volume integrals are correctly set up using the washer method. Which one is NOT?
Earlier you decided which curve was the outer radius for the region between $y=x$ and $y=x^2$.
The answer: $y = x$ is the outer radius on $[0,1]$ because $x \geq x^2$ there (e.g. at $x=0.5$: $0.5 > 0.25$). The volume is $\dfrac{2\pi}{15}$. Did your intuition match? The key insight is that "outer radius" simply means the curve farther from the axis of rotation — check with a test value whenever you're unsure.
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. Find the exact volume of the solid formed when the region bounded by $y = \sqrt{x}$ and $y = x^2$ is rotated about the $x$-axis. (2 marks)
Q2. The region enclosed by $y = 2x$ and $y = x^2$ is rotated about the $x$-axis. Find the exact volume. (3 marks)
Q3. Explain why the washer-method formula is $\pi\!\displaystyle\int_a^b[f(x)^2 - g(x)^2]\,dx$ and not $\pi\!\displaystyle\int_a^b[f(x)-g(x)]^2\,dx$. Support your explanation with a numerical example. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $y=x$ vs $y=x^3$: intersect at $x=0,1$. On $[0,1]$: $x \geq x^3$, so $y=x$ is outer.
2. $y=\sqrt{x}$ vs $y=x$: intersect at $x=0,1$. $\sqrt{x} \geq x$ on $(0,1)$. $V = \pi\!\int_0^1[x-x^2]\,dx$.
3. $\pi\!\int_0^1[x-x^4]\,dx = \pi\left[\tfrac{x^2}{2}-\tfrac{x^5}{5}\right]_0^1 = \pi(\tfrac{1}{2}-\tfrac{1}{5}) = \dfrac{3\pi}{10}$.
4. At $R=3, r=1$: $R^2-r^2 = 9-1=8$; $(R-r)^2=(3-1)^2=4$. These differ. The washer area is the area of the large circle minus the area of the small circle, not the square of the difference in radii.
5. $y=2x$ vs $y=x^2$: intersect at $x=0,2$. $R=2x$, $r=x^2$. $V = \pi\!\int_0^2[4x^2-x^4]\,dx = \pi\left[\tfrac{4x^3}{3}-\tfrac{x^5}{5}\right]_0^2 = \pi\!\left(\tfrac{32}{3}-\tfrac{32}{5}\right) = \pi \cdot 32 \cdot \tfrac{2}{15} = \dfrac{64\pi}{15}$.
Q1 (2 marks): $R=\sqrt{x}$, $r=x^2$, limits $0$ to $1$ [1]. $V = \pi\!\int_0^1[x-x^4]\,dx = \pi\!\left[\tfrac{x^2}{2}-\tfrac{x^5}{5}\right]_0^1 = \pi(\tfrac{1}{2}-\tfrac{1}{5}) = \mathbf{\dfrac{3\pi}{10}}$ cubic units [1].
Q2 (3 marks): Intersections: $2x=x^2 \Rightarrow x=0,2$ [1]. $R=2x$, $r=x^2$. $V=\pi\!\int_0^2[4x^2-x^4]\,dx = \pi\!\left[\tfrac{4x^3}{3}-\tfrac{x^5}{5}\right]_0^2 = \pi(\tfrac{32}{3}-\tfrac{32}{5})$ [1] $= \dfrac{64\pi}{15}$ cubic units [1].
Q3 (3 marks): Each washer (thin annular slice) has outer radius $R=f(x)$ and inner radius $r=g(x)$ [1]. Its cross-sectional area is $\pi R^2 - \pi r^2 = \pi(R^2-r^2)$, which is the area of the large disk minus the area of the hole — not $\pi(R-r)^2$ [1]. Numerical: at $R=2,r=1$: $\pi(4-1)=3\pi \neq \pi(2-1)^2=\pi$ [1].
Five timed questions on washer-method volume problems. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering volumes-between-curves questions. Lighter alternative to the boss.
Mark lesson as complete
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