1.8.3 Volume of Revolution - GCEALevelMaths9709

In this topic we will learn how to:

use definite integration to find a volume of revolution about one of the axes

We can use integration to find the volume produced by a graph when it rotates about an axis $360^{\circ}$ . This is known as the volume of revolution. To find this we use the two formulae below,
$V = \int \pi y^{2}\ dx$ $V = \int \pi x^{2}\ dy$ It is derived from the formula of volume of a cylinder, $V = \pi r^{2}\ h$ , hence the similarities. The first equation is used when we’re rotating about the $x$ -axis. The second equation is used when we’re rotating about the $y$ -axis.

Let’s apply this to some past paper questions.

1. The diagram shows part of the curve with equation $y = x^{2} + 1$ . The shaded region enclosed by the curve, the $y$ -axis and the line $y = 5$ is rotated through $360^{\circ}$ about the $y$ -axis. Find the volume obtained. (9709/12/F/M/20 number 3)

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The volume of revolution of this graph would look like this,

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$y = x^{2} + 1$ Since we’re rotating about the $y$ -axis we will use the formula,
$V = \int \pi x^{2}\ dy$ This means we need to find $x^{2}$ in terms of $y$ since we are working with respect to $y$ ,
$y = x^{2} + 1$ $\textcolor{#2192ff}{x^{2} = y - 1}$ Let’s substitute in $x^{2}$ ,
$V = \int \pi \textcolor{#2192ff}{x^{2}}\ dy$ $V = \int \pi \textcolor{#2192ff}{(y - 1)}\ dy$ We are already given the limits in terms of $y$ , so let’s substitute them in,
$V = \int_{1}^{5} \pi (y - 1)\ dy$ Note: If the limits are not in terms of $y$ and you’re rotating about the $y$ -axis, use the equation of the curve, to convert them to be in terms of $y$ .

Now let’s integrate,
$\int_{1}^{5} \pi (y - 1)\ dy$ $pi$ is a constant so you can move it outside the integral sign to make the integration easier,
$\pi \int_{1}^{5}(y - 1)\ dy$ $\pi \left[\frac{y^{2}}{2} - y\right]_{1}^{5}$ Substitute in the limits,
$\pi \left[\left(\frac{(5)^{2}}{2} - 5\right) - \left(\frac{(1)^{2}}{2} - 1\right)\right]$ $\pi \left[\frac{15}{2} - \left(-\frac{1}{2}\right)\right]$ $\pi \left[\frac{15}{2} + \frac{1}{2}\right]$ $8\pi$ Therefore, the final answer is,
$V = 8\pi$ 2. The diagram shows part of the curve with equation $y^{2} = x - 2$ and the lines $x = 5$ and $y = 1$ . The shaded region is enclosed by the curve and the lines is rotated through $360^{\circ}$ about the $x$ -axis. Find the volume obtained. (9709/12/M/J/21 number 9)

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The volume of revolution would look something like this,

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Notice that we only want the shaded region which is bounded by the curve and the line $y = 1$ . For that reason, we also need to consider the cylinder formed by the line $y = 1$ when it rotates,
$V = \int \pi y^{2}\ dx - \textmd{Area of cylinder}$ Let’s first find the volume produced by the curve,
$\int \pi \textcolor{#2192ff}{y^{2}}\ dx$ Substitute in $y^{2}$ ,
$\int \pi \textcolor{#2192ff}{(x - 2)}\ dx$ From the diagram, we can tell that one of our limits is $5$ . The other limit is at $y = 1$ , so we need to convert it to be in terms of $x$ ,
$\textmd{At } y = 1$ $y^{2} = x - 2$ $1^{2} = x - 2$ $1 = x - 2$ $x = 3$ Therefore, our limits are $3$ and $5$ ,
$\int_{3}^{5} \pi (x - 2)\ dx$ Now let’s integrate,
$\pi \int_{3}^{5} (x - 2)\ dx$ $\pi \left[\frac{x^{2}}{2} - 2x\right]_{3}^{5}$ Substitute in the limits,
$\pi \left[\left(\frac{(5)^{2}}{2} - 2(5)\right) - \left(\frac{(3)^{2}}{2} - 2(3)\right)\right]$ $\pi \left[\frac{5}{2} - \left(-\frac{3}{2}\right)\right]$ $\pi \left[\frac{5}{2} + \frac{3}{2}\right]$ $4\pi$ Now let’s find the volume produced by the cylinder due to the line $y = 1$ ,
$V = \pi r^{2}\ h$ $V = \pi \times 1^{2} \times (5 - 3)$ $V = 2\pi$ Therefore,
$V = 4\pi - 2\pi$ $V = 2\pi$ Therefore, the final answer is,
$V = 2\pi$

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