Volumes of Revolution Around the y-Axis
The same disc method works turned on its side — rearrange for x, swap the limits to y-values, and integrate exactly as before.
The disc method about the y-axis
When the region bounded by $x=g(y)$, the $y$-axis, and the lines $y=c$, $y=d$ is rotated $360°$ about the $y$-axis, the resulting solid has volume $$V=\pi\int_c^d x^2\,dy$$ This mirrors Theorem 14.1 exactly, with the roles of $x$ and $y$ swapped.
A curve is normally given as $y=f(x)$. To rotate about the $y$-axis, first rearrange to make $x$ the subject: $x=g(y)$. Then square and integrate with respect to $y$, using $y$-limits rather than $x$-limits.
Rearranging: $x^2=y$.
$V=\pi\displaystyle\int_0^4x^2\,dy=\pi\int_0^4y\,dy=\pi\left[\dfrac{y^2}{2}\right]_0^4=\pi\left(8-0\right)=8\pi$.
Rearranging: $y^2=x-1 \Rightarrow x=y^2+1$, so $x^2=(y^2+1)^2=y^4+2y^2+1$.
$V=\pi\displaystyle\int_0^3(y^4+2y^2+1)\,dy=\pi\left[\dfrac{y^5}{5}+\dfrac{2y^3}{3}+y\right]_0^3$.
$=\pi\left(\dfrac{243}{5}+\dfrac{54}{3}+3\right)=\pi\left(\dfrac{243}{5}+18+3\right)=\pi\left(\dfrac{243}{5}+21\right)=\pi\cdot\dfrac{348}{5}=\dfrac{348\pi}{5}$.
Rearranging: $x=e^y$, so $x^2=e^{2y}$.
$V=\pi\displaystyle\int_0^2e^{2y}\,dy=\pi\left[\dfrac{e^{2y}}{2}\right]_0^2=\pi\left(\dfrac{e^4}{2}-\dfrac12\right)=\dfrac{\pi(e^4-1)}{2}$.
The limits of integration for rotation about the $y$-axis are always values of $y$, not $x$ — using the original $x$-limits from the curve's description (rather than converting them to the corresponding $y$-values) is one of the most common errors in this topic.
If a question gives the region in terms of $x$-boundaries, always substitute into the original $y=f(x)$ equation to find the matching $y$-boundaries before setting up the integral — don't assume the numbers transfer directly.
Volumes of revolution for parametric curves
For a curve given parametrically by $x=f(t)$ and $y=g(t)$, the same substitution idea used for rotation about the $x$-axis applies here: since $dy=\dfrac{dy}{dt}\,dt$, substituting into $V=\pi\int x^2\,dy$ gives $$V=\pi\int_{t_1}^{t_2}x^2\,\dfrac{dy}{dt}\,dt$$ where $t_1$ and $t_2$ are the parameter values matching the original $y$-limits — you never need to eliminate $t$ and find a Cartesian equation first. See the x-axis lesson for the equivalent result when rotating about the $x$-axis.
$\dfrac{dy}{dt}=2t$, so $V=\pi\displaystyle\int x^2\dfrac{dy}{dt}\,dt=\pi\int(3t)^2(2t)\,dt=\pi\int18t^3\,dt$.
Limits: at $y=0$, $t^2=0\Rightarrow t=0$; at $y=4$, $t^2=4\Rightarrow t=2$ (taking $t\geq0$).
$V=\pi\displaystyle\int_0^218t^3\,dt=\pi\left[\dfrac{9t^4}{2}\right]_0^2=\pi\left(\dfrac{9(16)}{2}\right)=72\pi$.
Exam-style questions
Question 1 6 marks
The region bounded by $y=x^3$, the $y$-axis, and the lines $y=0,y=8$ (for $x\geq0$) is rotated $360°$ about the $y$-axis. Find the exact volume generated.
$x=y^{1/3}$, so $x^2=y^{2/3}$ (M1, A1)
$V=\pi\displaystyle\int_0^8y^{2/3}\,dy=\pi\left[\dfrac{3y^{5/3}}{5}\right]_0^8$ (M1, A1)
$8^{5/3}=(2^3)^{5/3}=2^5=32$, so $V=\pi\left(\dfrac{3(32)}{5}\right)=\dfrac{96\pi}{5}$ (M1, A1)
Question 2 6 marks
The region bounded by $x=y^2-2y+2$, the $y$-axis, and the lines $y=0,y=2$ is rotated $360°$ about the $y$-axis. Find the exact volume generated.
$x^2=(y^2-2y+2)^2=y^4-4y^3+8y^2-8y+4$ (M1, A1)
$V=\pi\displaystyle\int_0^2(y^4-4y^3+8y^2-8y+4)\,dy=\pi\left[\dfrac{y^5}{5}-y^4+\dfrac{8y^3}{3}-4y^2+4y\right]_0^2$ (M1, A1)
$=\pi\left(\dfrac{32}{5}-16+\dfrac{64}{3}-16+8\right)=\pi\left(\dfrac{32}{5}+\dfrac{64}{3}-24\right)$ (M1)
$=\pi\left(\dfrac{96}{15}+\dfrac{320}{15}-\dfrac{360}{15}\right)=\pi\cdot\dfrac{56}{15}=\dfrac{56\pi}{15}$ (A1)
Question 3 6 marks
The region bounded by $y=2x+1$, the $y$-axis, and the lines $y=1,y=5$ is rotated $360°$ about the $y$-axis. Find the exact volume generated, and verify it agrees with the volume of the equivalent cone/frustum found using standard geometric formulae.
$x=\dfrac{y-1}{2}$, so $x^2=\dfrac{(y-1)^2}{4}$ (M1, A1)
$V=\pi\displaystyle\int_1^5\dfrac{(y-1)^2}{4}\,dy=\dfrac{\pi}{4}\left[\dfrac{(y-1)^3}{3}\right]_1^5=\dfrac{\pi}{4}\left(\dfrac{64}{3}-0\right)=\dfrac{16\pi}{3}$ (M1, A1, A1)
This is a cone-shaped solid, radius $2$ at $y=5$, apex at $y=1$, height $4$: $V=\dfrac13\pi r^2h=\dfrac13\pi(2)^2(4)=\dfrac{16\pi}{3}$ ✓ (B1)