Volumes of Revolution Around the y-Axis
Same formula as Year 1 — but now every integration technique in the book is fair game for evaluating it.
Revisiting the volume of revolution formula
Recall that rotating the region between $x=g(y)$ and the $y$-axis, between $y=c$ and $y=d$, fully around the $y$-axis creates a solid of volume
$$V=\pi\int_c^dx^2\,dy$$
What's new in Year 2 is the difficulty of the curves involved — you'll now need every integration technique from this book (trigonometric identities, inverse trig results, partial fractions, integration by substitution) to actually evaluate $\displaystyle\int x^2\,dy$.
$V=\pi\displaystyle\int_1^3x^2\,dy=\pi\int_1^3\frac1{y(y+1)}\,dy$.
Partial fractions: $\dfrac1{y(y+1)}=\dfrac1y-\dfrac1{y+1}$ (a standard result from the method of differences chapter).
$V=\pi\Big[\ln|y|-\ln|y+1|\Big]_1^3=\pi\Big[\ln\left(\dfrac y{y+1}\right)\Big]_1^3=\pi\left(\ln\dfrac34-\ln\dfrac12\right)=\pi\ln\dfrac{3/4}{1/2}=\pi\ln\dfrac32$.
Before integrating, always find $x^2$ explicitly and fully simplify it — it's easy to make a sign or index error squaring an expression involving fractions, roots or trig functions. Writing $x^2=\ldots$ as its own line of working, before substituting into $V=\pi\displaystyle\int x^2\,dy$, avoids this.
A worked example needing a trig identity
$V=\pi\displaystyle\int_0^{\pi/4}\sin^22y\,dy$. Using $\sin^2\theta=\dfrac12(1-\cos2\theta)$ with $\theta=2y$: $\sin^22y=\dfrac12(1-\cos4y)$.
$V=\pi\displaystyle\int_0^{\pi/4}\frac12(1-\cos4y)\,dy=\frac\pi2\left[y-\frac14\sin4y\right]_0^{\pi/4}$
$=\dfrac\pi2\left(\left(\dfrac\pi4-\dfrac14\sin\pi\right)-(0-0)\right)=\dfrac\pi2\left(\dfrac\pi4-0\right)=\dfrac{\pi^2}{8}$.
Exam-style questions
Question 1 7 marks
The region $R$ is bounded by the curve $x=\dfrac1{\sqrt{(y+1)(y+2)}}$, the $y$-axis, and the lines $y=0$ and $y=2$. Find the exact volume generated when $R$ is rotated $2\pi$ radians about the $y$-axis.
$V=\pi\displaystyle\int_0^2x^2\,dy=\pi\int_0^2\frac1{(y+1)(y+2)}\,dy$. (M1, A1)
Partial fractions: $\dfrac1{(y+1)(y+2)}=\dfrac1{y+1}-\dfrac1{y+2}$. (M1, A1)
$V=\pi\Big[\ln|y+1|-\ln|y+2|\Big]_0^2=\pi\left(\left(\ln3-\ln4\right)-\left(\ln1-\ln2\right)\right)=\pi\ln\dfrac{3/4}{1/2}=\pi\ln\dfrac32$. (M1, A1, A1)
Question 2 8 marks
The region $R$ is bounded by $x=\dfrac1{\sqrt{4+y^2}}$, the $y$-axis and the lines $y=0$ and $y=2\sqrt3$. Find the exact volume generated when $R$ is rotated $2\pi$ radians about the $y$-axis, leaving your answer in terms of $\pi$.
$x^2=\dfrac1{4+y^2}$, so $V=\pi\displaystyle\int_0^{2\sqrt3}\dfrac1{4+y^2}\,dy$; this matches the standard $\arctan$ form with $a=2$. (M1, A1)
$=\pi\left[\dfrac12\arctan\left(\dfrac y2\right)\right]_0^{2\sqrt3}=\dfrac\pi2\Big(\arctan\sqrt3-\arctan0\Big)$. (M1, A1)
$=\dfrac\pi2\times\dfrac\pi3=\dfrac{\pi^2}{6}$. (A1, A1)