Category Archives: Year 12 Specialist Mathematics

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

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December 28, 2024 · 7:12 am

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.

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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.

This is a better view of the 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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