Solids of Revolution — About the y-axis
You know how to spin a curve about the x-axis to get a solid. But what if the axis of rotation is vertical? The same disk logic applies — you just integrate with respect to y. In this lesson you'll rewrite curves as $x = f(y)$, build disks of radius $f(y)$, and integrate vertically: $V = \pi\displaystyle\int_c^d [f(y)]^2\,dy$.
Spin $y = x^2$ about the y-axis from $y = 0$ to $y = 4$ (i.e. $x$ from 0 to 2). Without using a formula — do you expect a wider or narrower solid than rotating about the x-axis over the same interval? Write your reasoning below.
In Lesson 4 you integrated horizontally — summing disks stacked along the x-axis. Rotating about the y-axis means the disks are stacked vertically. Each horizontal disk at height $y$ has radius equal to the x-value of the curve at that height.
To use the disk formula you need $x$ as a function of $y$. If the original equation is $y = f(x)$, rearrange it to get $x = g(y)$. Then each disk at height $y$ has:
- Radius $= g(y)$ (the $x$-value of the curve)
- Cross-sectional area $= \pi\,[g(y)]^2$
- Thickness $= dy$ (an infinitesimally thin horizontal slice)
Summing from $y = c$ to $y = d$:
$V = \pi\displaystyle\int_c^d [g(y)]^2\,dy$
Key facts
- Disk formula about the $y$-axis: $V = \pi\displaystyle\int_c^d [g(y)]^2\,dy$
- The limits of integration are $y$-values, not $x$-values
- The integrand $[g(y)]^2$ is the square of the x-expression
Concepts
- Why rotating about the $y$-axis requires $x$ as a function of $y$
- How horizontal disk slices generate the solid
- The relationship between the curve shape and the size of each disk
Skills
- Rearrange a curve equation to express $x$ in terms of $y$
- Identify correct $y$-limits from context or from an $x$-interval
- Evaluate $V = \pi\displaystyle\int_c^d [g(y)]^2\,dy$ for polynomial and radical curves
When a region bounded by $x = g(y)$, the $y$-axis, and horizontal lines $y = c$, $y = d$ is rotated about the $y$-axis, the volume of the resulting solid is:
Where does this come from? Slice the solid into thin horizontal disks. Each disk at height $y$ has:
- Radius $r = g(y)$ (the horizontal reach of the curve from the $y$-axis)
- Cross-sectional area $A = \pi r^2 = \pi[g(y)]^2$
- Volume of thin slice $\approx A \cdot \Delta y = \pi[g(y)]^2\,\Delta y$
Taking the limit as $\Delta y \to 0$ and summing from $c$ to $d$ gives the integral formula.
Standard approach:
- Rearrange $y = f(x)$ to get $x = g(y)$.
- Convert $x$-limits to $y$-limits: substitute the $x$-endpoints into $y = f(x)$.
- Set up the integral $V = \pi\displaystyle\int_c^d [g(y)]^2\,dy$.
- Expand $[g(y)]^2$, integrate term by term, and evaluate.
When a region bounded by $x = g(y)$, the $y$-axis, and horizontal lines $y = c$, $y = d$ is rotated about the $y$-axis, the volume of the resulting solid is:
Pause — copy the $y$-axis disk method formula $V=\pi\int_c^d[g(y)]^2\,dy$ and the rearrangement rule: solve the original curve equation for $x$ in terms of $y$ into your book.
Quick check: The curve $y = x^3$ is rotated about the $y$-axis from $x = 0$ to $x = 2$. Which integral gives the volume?
We just saw that rotating about the $y$-axis uses $V=\pi\int_c^d[g(y)]^2\,dy$ where $x=g(y)$ is the curve rearranged in terms of $y$. That raises a question: when converting from $x$-limits to $y$-limits, what is the systematic procedure, and what happens if you forget to rearrange? This card answers it → substitute $x=a,b$ into $x=g(y)$ to get $y=c,d$; failing to rearrange leaves $y$ as the variable of integration with an $x$-expression, which is meaningless.
The two most important mechanical steps for the y-axis method are (1) rearranging the curve and (2) finding the new limits.
Step 1 — Rearrange $y = f(x)$ to $x = g(y)$:
- $y = x^2 \Rightarrow x = \sqrt{y}$ (positive root if $x \geq 0$)
- $y = x^3 \Rightarrow x = y^{1/3}$
- $y = 4x \Rightarrow x = y/4$
- $y = \sqrt{x} \Rightarrow x = y^2$
Step 2 — Convert $x$-limits to $y$-limits:
If $x$ runs from $a$ to $b$, substitute into $y = f(x)$:
$c = f(a)$ and $d = f(b)$.
Example: Rotate $y = x^2$ about the y-axis from $x = 0$ to $x = 3$.
- Rearrange: $x = \sqrt{y}$ so $[g(y)]^2 = y$.
- Limits: $y = 0^2 = 0$ and $y = 3^2 = 9$.
- Integral: $V = \pi\displaystyle\int_0^9 y\,dy = \pi\Big[\dfrac{y^2}{2}\Big]_0^9 = \pi \cdot \dfrac{81}{2} = \dfrac{81\pi}{2}$ units$^3$.
The two most important mechanical steps for the y -axis method are (1) rearranging the curve and (2) finding the new limits.
Pause — copy the limit-conversion steps: substitute $x$-endpoints into the rearranged equation to get $y$-limits; include a worked example into your book.
Did you get this? True or false: to rotate $y = x^2$ about the $y$-axis from $x = 0$ to $x = 3$, the correct $y$-limits are $c = 0$ and $d = 9$.
Worked examples · 3 in a row, reveal as you go
Find the volume of the solid formed when the region bounded by $y = x^2$, the $y$-axis, and $y = 4$ is rotated about the $y$-axis.
The region enclosed by $y = x^3$, the $y$-axis, $x = 0$ and $x = 2$ is rotated about the $y$-axis. Find the exact volume.
Find the volume of the cone formed when the line $y = 3x$ (for $x \geq 0$) is rotated about the $y$-axis from $y = 0$ to $y = 6$.
Fill the gap: The curve $y = \sqrt{x}$ rearranged as $x = g(y)$ gives $x = $ . Then $[g(y)]^2 = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when rotating $y = x^2$ about the $y$-axis from $x = 1$ to $x = 3$, the $y$-limits for the integral are $c = 1$ and $d = 9$.
Activities · practice with the ideas
Find the volume when $y = x^2$ is rotated about the $y$-axis from $x = 0$ to $x = 2$.
The region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 9$ is rotated about the $y$-axis. Find the volume. Hint: rearrange to $x = y^2$.
Find the volume of the solid formed by rotating $y = 2x$ about the $y$-axis from $y = 0$ to $y = 4$ and verify using the cone formula $V = \frac{1}{3}\pi r^2 h$.
The curve $x = y^2 - 1$ (for $y \geq 0$) is rotated about the $y$-axis from $y = 1$ to $y = 3$. Set up and evaluate the volume integral. Note: $g(y) = y^2 - 1$ is already expressed in terms of y.
Show that rotating $y = x^n$ (for $x \geq 0$) about the $y$-axis from $x = 0$ to $x = a$ gives $V = \dfrac{\pi n}{n+2}\,a^2 \cdot a^n / a^{n-2}$… Actually: derive a general formula for the volume in terms of $n$ and $a$.
Odd one out: Three of these statements about rotating $y = x^2$ (for $x \geq 0$) about the $y$-axis are correct. Which one is NOT?
Earlier you estimated whether rotating $y = x^2$ about the $y$-axis gives a larger or smaller solid than rotating about the $x$-axis over the same $x$-interval.
About the $y$-axis (from $x = 0$ to $x = 2$, i.e. $y = 0$ to $y = 4$): $V = 8\pi$.
About the $x$-axis (from $x = 0$ to $x = 2$): $V = \pi\displaystyle\int_0^2 x^4\,dx = \pi\cdot\dfrac{32}{5} = \dfrac{32\pi}{5} \approx 6.4\pi$.
The $y$-axis solid is larger ($8\pi > 6.4\pi$). Did your intuition get it right?
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 volume of the solid formed when the region bounded by $y = x^2$, $x = 0$, and $y = 9$ is rotated about the $y$-axis. (2 marks)
Q2. The region bounded by $y = x^3$, $x = 0$, and $x = 2$ is rotated about the $y$-axis. Find the exact volume. (3 marks)
Q3. A solid is formed by rotating $x = \sqrt{4 - y}$ about the $y$-axis from $y = 0$ to $y = 4$. Find the volume and describe the shape. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $V = \pi\displaystyle\int_0^4 y\,dy = 8\pi$ · 2. $x = y^2$, $[g(y)]^2 = y^4$, $V = \pi\displaystyle\int_0^3 y^4\,dy = \pi\Big[\dfrac{y^5}{5}\Big]_0^3 = \dfrac{243\pi}{5}$ · 3. $x = y/2$, $[g(y)]^2 = y^2/4$, $V = \dfrac{\pi}{4}\displaystyle\int_0^4 y^2\,dy = \dfrac{\pi}{4}\cdot\dfrac{64}{3} = \dfrac{16\pi}{3}$; cone: $r=2, h=4 \Rightarrow V = \dfrac{16\pi}{3}$ ✓ · 4. $\displaystyle\int_1^3(y^4-2y^2+1)\,dy = \Big[\dfrac{y^5}{5}-\dfrac{2y^3}{3}+y\Big]_1^3 = \Big(\dfrac{243}{5}-18+3\Big)-\Big(\dfrac{1}{5}-\dfrac{2}{3}+1\Big) = \dfrac{242}{5}+\dfrac{2}{3}-16 = \dfrac{726+10-240}{15} = \dfrac{496}{15}$; $V = \dfrac{496\pi}{15}$ · 5. $x = y^{1/n}$, $[g(y)]^2 = y^{2/n}$, $y$-limits $0$ to $a^n$; $V = \pi\displaystyle\int_0^{a^n} y^{2/n}\,dy = \pi\cdot\dfrac{n}{n+2}(a^n)^{(n+2)/n} = \dfrac{\pi n a^{n+2}}{n+2}$.
Q1 (2 marks): $x = \sqrt{y}$, so $[g(y)]^2 = y$ [1]. $V = \pi\displaystyle\int_0^9 y\,dy = \pi\Big[\dfrac{y^2}{2}\Big]_0^9 = \dfrac{81\pi}{2}$ units$^3$ [1].
Q2 (3 marks): $x = y^{1/3}$, $[g(y)]^2 = y^{2/3}$ [1]; limits $y = 0$ to $8$ [1]; $V = \pi\displaystyle\int_0^8 y^{2/3}\,dy = \pi\cdot\dfrac{3}{5}\Big[y^{5/3}\Big]_0^8 = \dfrac{3\pi}{5}\cdot 32 = \dfrac{96\pi}{5}$ units$^3$ [1].
Q3 (3 marks): $[g(y)]^2 = (\sqrt{4-y})^2 = 4-y$ [1]; $V = \pi\displaystyle\int_0^4(4-y)\,dy = \pi\Big[4y - \dfrac{y^2}{2}\Big]_0^4 = \pi(16-8) = 8\pi$ units$^3$ [1]; this is a paraboloid of revolution — a paraboloid that tapers to a point at the top and is widest at the base [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 y-axis rotation questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.