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December 30, 2024 · 8:16 am

Volume of revolution about a line that is not an axis

Find the volume of the solid of revolution obtained by rotating the region bounded by $f(x)=x^3+1, g(x)=x^2, 0\le x\le 1$ about the line $y=3$ .

Rotate the green region about the line $y=3$

Washer Method

$\begin{equation*}V=\pi \int [f(x)]^2 dx \end{equation}$

The volume of the solid is the volume of $y=x^2$ rotated about $y=3$ subtract the volume of $y=x^3+1$ rotated about $y=3$ .

$\begin{equation*}V=\pi \int_0^1((3-x^2)^2-(3-(x^3+1))^2 dx\end{equation}$

$3-x^2$ is the distance (i.e radius) of the curve and the line.

$\begin{equation*}V=\pi \int_0^1(9-6x^2+x^4-(4-4x^3+x^6)) dx\end{equation}$

$\begin{equation*}V=\pi \int_0^1(5-6x^2+4x^3+x^4-x^6) dx\end{equation}$

$\begin{equation*}V=\pi (5x-2x^3+x^4+\frac{x^5}{5}-\frac{x^7}{7}]_0^1\end{equation}$

$\begin{equation*}V=\pi (5-2+1+\frac{1}{5}-\frac{1}{7})\end{equation}$

$\begin{equation*}V=\frac{142 \pi}{35}\end{equation}$

Shell Method

The shell method is much harder because we need to split the integral into two parts.

We need to rotate the green region about $y=3$ and the red region

$\begin{equation*}V=2\pi\int (xf(x))dx\end{equation}$

$\begin{equation*}V=2\pi[\int_0^1(3-y)\sqrt{y} dy+\int_1^2 (3-y)(y-1)^{\frac{1}{3}} dy]\end{equation}$

$3-y$ is the distance between each $y-$ value and the line of rotation. For example, if we were rotating about the $x-$ axis, the distance is $y$ .

$\sqrt{y}$ is the height of the cylinder between $0$ and $1$ . $1-(y-1)^{\frac{1}{3}}$ is the height of the cylinder between $1$ and $2$ . Refer back to Shell method for more information.

I used a calculator to find this integral

Volume of Revolution – Method One (Disc or Washer Method)

If we rotate this line segment around the $x-$ axis, we generate a three dimensional solid.

We are going to find the volume of this solid.

Consider a small section of the line segment and rotate this about the $x-$ axis.

As the width of the section $(\delta x)$ gets smaller (i.e. $\rightarrow 0$ ), the solid is a cylinder.

The radius of the cylinder is $f(x)$ and the height of the cylinder is $\delta x$ .

The volume of a cylinder is $V=\pi r^2 h$

Hence the volume of our section is

$\begin{equation*}V=\pi[f(x)]^2\delta x\end{equation}$

If we divide our line segment into a large number of cylinders (of equal height) then,

$\begin{equation*}V=\Sigma_a^b(\pi [f(x)]^2\delta x\end{equation}$

where $a$ is the lower $x$ value and $b$ the upper.

Now we want $\delta x\rightarrow 0$ so $V=\lim\limits_{\delta x \to 0} \Sigma_a^b(\pi [f(x)]^2\delta x$

Which is

$\begin{equation*}V=\int_a^b \pi [f(x)]^2 dx\end{equation}$

Example

The curve $y=\sqrt{x-1}$ , where $2\le x\le5$ is rotated about the $x-$ axis to form a solid of revolution. Find the volume of this solid.

$\begin{equation*}V=\pi \int_2^5( y^2 dx)\end{equation}$

$\begin{equation*}V=\pi \int_2^5 x-1 \space dx \end{equation}$

$\begin{equation*}V=\pi (\frac{x^2}{2}-x]_2^5)\end{equation}$

$\begin{equation*}V=\pi(\frac{25}{2}-5-(\frac{4}{2}-2))\end{equation}$

$\begin{equation*}V=\frac{15 \pi}{2}\end{equation}$

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Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

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Tag Archives: Disc method

Volume of revolution about a line that is not an axis

Volume of Revolution – Method One (Disc or Washer Method)

Example

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