Question 17
The region between the $x$-axis and the curve of the function $y=1+\sin(2x)$ for $0\le x\le\dfrac{\pi}{2}$ is rotated about the $x$-axis to form a solid of revolution.
Determine the volume of this solid. Express your answer in simplest form.
[5 marks]Spin the region and slice it into discs
The region turns about the $x$-axis and sweeps out the solid. Each slice is a disc, and adding up the discs gives the volume integral.
The curve never drops below $y=1$, so every slice is a solid disc with no hole in the middle.
Build the area under the squared curve from its three pieces
The volume is $\pi$ times the area under the graph of $y^2$. Expanding $y^2$ splits that area into three pieces. Turn the pieces on and off to see what each one adds, and slide the disc along the solid.
All three pieces together fill the area under the graph of $y^2$ exactly. Multiply by $\pi$ and you have the volume, $V=\pi\left(\frac{3\pi}{4}+2\right)$.
This is what you get if you square each part on its own. $(1+\sin(2x))^2$ is not $1+\sin^2(2x)$, because the middle term $2\sin(2x)$ is missing. The shading falls short of the curve by an area of $2$.
The identity splits $\sin^2(2x)$ into the constant $\frac12$ and the term $-\frac12\cos(4x)$. Over this interval the cosine part has equal areas above and below its axis, so they cancel. The area is $\frac12\times\frac{\pi}{2}=\frac{\pi}{4}$.
Only some of the pieces are shaded, so the shading does not reach the black curve. Turn on all three pieces of the expansion.
Nothing is shaded. Turn on the pieces of the expansion to fill the area under the graph of $y^2$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA sample response.
For a region rotated about the $x$-axis, each slice is a disc of radius $y$, so the volume is $\pi$ times the integral of $y^2$. Substitute the function and the limits from the question.
$$V=\pi\int_a^b y^2\,dx$$
$$V=\pi\int_0^{\frac{\pi}{2}}\big(1+\sin(2x)\big)^2\,dx$$
Marker: correctly substitutes into the appropriate volume of a solid of revolution rule
You cannot integrate the square of a bracket directly, so expand it with $(a+b)^2=a^2+2ab+b^2$. The middle term is the one people drop.
$$\begin{aligned}V&=\pi\int_0^{\frac{\pi}{2}}1+2\sin(2x)\\&\qquad+\sin^2(2x)\,dx\end{aligned}$$
If something went wrong in the first line, you can still earn this mark and the later ones by expanding and working correctly from what you wrote.
Marker: expands an integrand
There is no rule for integrating $\sin^2$ directly. The double-angle identity $\cos(2A)=1-2\sin^2(A)$ rearranges to $\sin^2(A)=\frac12\big(1-\cos(2A)\big)$. Here $A=2x$, so the angle doubles to $4x$.
$$\begin{aligned}V&=\pi\int_0^{\frac{\pi}{2}}1+2\sin(2x)\\&\qquad+\frac12\big(1-\cos(4x)\big)\,dx\end{aligned}$$
$$\begin{aligned}V&=\pi\int_0^{\frac{\pi}{2}}\Big(\frac32+2\sin(2x)\\&\qquad-\frac12\cos(4x)\Big)\,dx\end{aligned}$$
Marker: uses a suitable double-angle identity to enable an integration process to be completed
Integrate term by term. For $\sin(kx)$ and $\cos(kx)$ you divide by $k$, so $2\sin(2x)$ becomes $-\cos(2x)$ and $-\frac12\cos(4x)$ becomes $-\frac18\sin(4x)$.
$$\begin{aligned}V&=\pi\Big(\frac32x-\cos(2x)\\&\qquad-\frac18\sin(4x)\Big)\Bigg|_0^{\frac{\pi}{2}}\end{aligned}$$
An equivalent integrated expression is accepted, such as $\pi\left(x-\cos(2x)+\frac{x}{2}-\frac18\sin(4x)\right)$.
Marker: integrates an expression
Substitute both limits. Use exact values: $\cos(\pi)=-1$, $\sin(2\pi)=0$, $\cos(0)=1$ and $\sin(0)=0$. The bracket at the lower limit is not zero, so keep it.
$$\begin{aligned}V&=\pi\Big[\Big(\frac{3\pi}{4}-\cos(\pi)\\&\qquad\quad-\frac18\sin(2\pi)\Big)\\&\qquad-\big(0-\cos(0)-0\big)\Big]\\&=\pi\Big[\Big(\frac{3\pi}{4}+1\Big)-(-1)\Big]\end{aligned}$$
An equivalent volume is accepted, such as $\frac{3\pi^2}{4}+2\pi$.
Marker: determines volume in simplest form
Each phrase of the question gave you something
Setting up the volume integral is routine, but the question then leaves you to work out how to integrate it. You have to expand the square, notice that $\sin^2(2x)$ has no direct antiderivative, choose the double-angle identity with the angle doubled to $4x$, and evaluate exact trigonometric values at both limits. Nothing in the wording points you to the identity, so you join those steps yourself. It is still complex familiar because “rotated about the $x$-axis to form a solid of revolution” tells you exactly which rule to start from.
If I had to guess where you might lose marks, it would be in the expansion and the identity. Writing $1+\sin^2(2x)$ for the square loses the $2\sin(2x)$ term, and using $\cos(2x)$ instead of $\cos(4x)$ in the identity spoils the integration that follows. After that, watch the lower limit. The bracket there is $0-\cos(0)-0=-1$, and subtracting it adds $1$. That is where the second $1$ in the $+2$ comes from. The marking guide gives the last mark for the volume in simplest form, so finish with $\pi\left(\frac{3\pi}{4}+2\right)$ or $\frac{3\pi^2}{4}+2\pi$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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