Volumes of Revolution of Parametrically Defined Curves
Swap $dx$ for $\dfrac{dx}{dt}\,dt$ and convert the limits — the rest of the method is exactly as before.
Adjusting the formula for parametric curves
If a curve is given parametrically by $x=x(t)$, $y=y(t)$, the volume formula $V=\pi\displaystyle\int y^2\,dx$ is adapted by substituting $dx=\dfrac{dx}{dt}\,dt$ and converting the $x$-limits into $t$-limits:
$$V=\pi\int_{t_1}^{t_2}y^2\,\frac{dx}{dt}\,dt$$
where $t_1$ and $t_2$ are the parameter values corresponding to $x=a$ and $x=b$. (The equivalent $y$-axis version substitutes $dy=\dfrac{dy}{dt}\,dt$ instead.)
- Find $y^2$ in terms of $t$.
- Find $\dfrac{dx}{dt}$.
- Convert the $x$-limits into $t$-limits by solving $x(t)=a$ and $x(t)=b$.
- Substitute everything into $V=\pi\displaystyle\int_{t_1}^{t_2}y^2\dfrac{dx}{dt}\,dt$ and integrate with respect to $t$.
$y=2t\Rightarrow y^2=4t^2$. $\dfrac{dx}{dt}=2t$.
Limits: $x=0\Rightarrow t=0$; $x=4\Rightarrow t^2=4\Rightarrow t=2$ (taking $t\geq0$).
$V=\pi\displaystyle\int_0^24t^2\times2t\,dt=\pi\int_0^28t^3\,dt=\pi\Big[2t^4\Big]_0^2=\pi(32-0)=32\pi$.
Just as with Cartesian volumes, the resulting integral in $t$ can require any integration technique — including partial fractions, if $y^2\dfrac{dx}{dt}$ turns out to be a rational function of $t$ with a factorisable denominator.
$y^2=\dfrac1{(2+t)^2}$. $\dfrac{dx}{dt}=2+2t$.
Limits: $x=0\Rightarrow t(2+t)=0\Rightarrow t=0$ (rejecting $t=-2$ since $t>0$); $x=3\Rightarrow t^2+2t-3=0\Rightarrow(t+3)(t-1)=0\Rightarrow t=1$ (rejecting $t=-3$).
$V=\pi\displaystyle\int_0^1\frac1{(2+t)^2}\times(2+2t)\,dt=\pi\int_0^1\frac{2+2t}{(2+t)^2}\,dt$.
Partial fractions: $\dfrac{2+2t}{(2+t)^2}=\dfrac A{2+t}+\dfrac B{(2+t)^2}$; $2+2t=A(2+t)+B$. Setting $t=-2$: $-2=B$. Comparing $t$ coefficients: $A=2$.
$V=\pi\displaystyle\int_0^1\left(\frac2{2+t}-\frac2{(2+t)^2}\right)dt=\pi\Big[2\ln|2+t|+2(2+t)^{-1}\Big]_0^1$
$=\pi\left(\left(2\ln3+\dfrac23\right)-\left(2\ln2+1\right)\right)=\pi\left(2\ln\dfrac32-\dfrac13\right)$.
Exam-style questions
Question 1 8 marks
A curve has parametric equations $x=3t$, $y=\dfrac1{\sqrt t}$, for $t>0$. The region $R$ is bounded by the curve, the $x$-axis, and the lines $x=3$ and $x=6$. Find the exact volume generated when $R$ is rotated $2\pi$ radians about the $x$-axis.
$y^2=\dfrac1t$; $\dfrac{dx}{dt}=3$. (M1, A1)
Limits: $x=3\Rightarrow t=1$; $x=6\Rightarrow t=2$. (M1, A1)
$V=\pi\displaystyle\int_1^2\dfrac1t\times3\,dt=3\pi\Big[\ln t\Big]_1^2=3\pi(\ln2-0)=3\pi\ln2$. (M1, A1, A1, A1)