Tag Archives: Volume of revolution

January 1, 2025 · 9:40 am

Volume of Rotation About a Slanting Line

Given the area in the first quadrant bounded by $x^2=12y$ , the line $y=3$ and the $y-$ axis. What is the volume generated when this area is rotated about the line $2x-y+4=0$ ?

Rotate the green region about the line $y=2x+4$

Rotate the green region about the line $y=2x+4$

We can split the solid into shells.

$\begin{equation*}V=2\pi r dx dy\end{equation}$

Where $r$ is the distance from each $(x,y)$ point in the region to the line $2x-y+4=0$ , $dx$ is the width, and $dy$ is the height.

The distance between a point and a line is

$\begin{equation*} d=\frac{Ax+By+C=0}{\sqrt{A^2+B^2}}\end{equation}$

Hence, $r=\frac{2x-y+4}{\sqrt{5}}$

$\begin{equation*}V=2\pi\int \int \frac{2x-y+4}{\sqrt{5}} dx dy\end{equation}$

Now we just need to work out the bounds.

$0\le y \le 3$ and $0\le x\le \sqrt{12y}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}} \int_0^3 \int_0^{\sqrt{12y}} 2x-y+4 dx dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 x^2-yx+4x]_0^{\sqrt{12y}} dy \end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 12y-\sqrt{12}y^{\frac{3}{2}+4\sqrt{12y} dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(6y^2-\frac{2\sqrt{12}}{5}y^{\frac{5}{2}}+\frac{8\sqrt{12}}{3}y^{\frac{3}{2}}]_0^3\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{2\sqrt{12}}{5}(9\sqrt{3})+\frac{8\sqrt{12}}{3}(3\sqrt{3}))\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{108}{5}+48)\end{equation}$

$\begin{equation*}V=\frac{804\pi}{\sqrt{5}}\end{equation}$

If we rationalise the denominator

$\begin{equation*}V=\frac{804\sqrt{5}\pi}{25}\end{equation}$

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 Two (Shell Method)

I am going to use the same example as I did for Method One (Disc or Washer Method).

If we rotate the shaded region about the $x-$ axis, we get an open hollow cylinder (like a pipe).

The width of the integral is $\delta y$ and the midpoint is $y$ .

The height of the cylinder is $x$ , but we need it in terms of $y$ , hence $x=f(y)$

The volume of the hollow cylinder is the volume of the outer cylinder subtract the volume of the inner cylinder.

$\begin{equation*}V=\pi (y+\frac{\delta y}{2})^2f(y)-\pi (y-\frac{\delta y}{2})^2 f(y)\end{equation}$

$\begin{equation*}V=\pi f(y)((y+\frac{\delta y}{2})^2-(y-\frac{\delta y}{2})^2)\end{equation}$

Which we can expand using a difference of squares.

$\begin{equation*}V=\pi f(y)(y+\frac{\delta y}{2}+y-\frac{\delta y}{2})(y+\frac{\delta y}{2}-y+\frac{\delta y}{2})\end{equation}$

$\begin{equation*}V=\pi f(y)(2y \delta y)\end{equation}$

$\begin{equation*}V=2\pi yf(y)\delta y\end{equation}$

The volume of the entire sold will be

$\begin{equation*}V=\Sigma_{y=a}^b 2 \pi yf(y)\delta y\end{equation}$

As $\delta y \rightarrow 0$

$\begin{equation*}V=\lim\limits_{\delta y \to 0}\Sigma_{y=a}^b 2 \pi yf(y)\delta y=\int_a^b 2\pi yf(y) dy\end{equation}$

Even though we are rotating the line about the $x-$ axis, we are integrating with respect to the $y-$ axis.

Example

Find the volume of the solid generated by revolving the region between $y=x^2$ and $y=2x$ about the $y-$ axis.

If we are rotating about the $y-$ axis, we will integrate with respect to $x$ .

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

The height of our hollow cylinder is $2x-x^2$

Hence

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

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

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

$\begin{equation*}V=2\pi (\frac{2}{3}\times 8-\frac{1}{4}\times 16 )\end{equation}$

$\begin{equation*}V=32 \pi(\frac{1}{3}-\frac{1}{4})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

Let’s check with method one.

$x^2=y$ and $x=\frac{y}{2}$

$\begin{equation*}V=\pi \int_0^4 y-\frac{y^2}{4} dy\end{equation}$

$\begin{equation*}V=\pi (\frac{y^2}{2}-\frac{y^3}{12})]_0^4\end{equation}$

$\begin{equation*}V=\pi(8-\frac{16}{3})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

I try to pick the method that makes the integration easier.

1 Comment

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

Tagged as shell method, Volume of revolution

December 23, 2024 · 3:45 am

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.

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

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

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

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

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

1 Comment

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

Tagged as Disc method, Integration, volume, Volume of revolution

Tag Archives: Volume of revolution

Volume of Rotation About a Slanting Line

Volume of revolution about a line that is not an axis

Volume of Revolution Method Two (Shell Method)

Example

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

Example

Recent Posts

Categories

Archives