1.8.3 Volume of Revolution

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 360360^{\circ}. This is known as the volume of revolution. To find this we use the two formulae below,
V=πy2 dxV = \int \pi y^{2}\ dxV=πx2 dyV = \int \pi x^{2}\ dyIt is derived from the formula of volume of a cylinder, V=πr2 hV = \pi r^{2}\ h, hence the similarities. The first equation is used when we’re rotating about the xx-axis. The second equation is used when we’re rotating about the yy-axis.

Let’s apply this to some past paper questions.

1. The diagram shows part of the curve with equation y=x2+1y = x^{2} + 1. The shaded region enclosed by the curve, the yy-axis and the line y=5y = 5 is rotated through 360360^{\circ} about the yy-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=x2+1y = x^{2} + 1Since we’re rotating about the yy-axis we will use the formula,
V=πx2 dyV = \int \pi x^{2}\ dyThis means we need to find x2x^{2} in terms of yy since we are working with respect to yy,
y=x2+1y = x^{2} + 1x2=y1\textcolor{#2192ff}{x^{2} = y - 1}Let’s substitute in x2x^{2},
V=πx2 dyV = \int \pi \textcolor{#2192ff}{x^{2}}\ dyV=π(y1) dyV = \int \pi \textcolor{#2192ff}{(y - 1)}\ dyWe are already given the limits in terms of yy, so let’s substitute them in,
V=15π(y1) dyV = \int_{1}^{5} \pi (y - 1)\ dyNote: If the limits are not in terms of yy and you’re rotating about the yy-axis, use the equation of the curve, to convert them to be in terms of yy.

Now let’s integrate,
15π(y1) dy\int_{1}^{5} \pi (y - 1)\ dypipi is a constant so you can move it outside the integral sign to make the integration easier,
π15(y1) dy\pi \int_{1}^{5}(y - 1)\ dyπ[y22y]15\pi \left[\frac{y^{2}}{2} - y\right]_{1}^{5}Substitute in the limits,
π[((5)225)((1)221)]\pi \left[\left(\frac{(5)^{2}}{2} - 5\right) - \left(\frac{(1)^{2}}{2} - 1\right)\right]π[152(12)]\pi \left[\frac{15}{2} - \left(-\frac{1}{2}\right)\right]π[152+12]\pi \left[\frac{15}{2} + \frac{1}{2}\right]8π8\piTherefore, the final answer is,
V=8πV = 8\pi2. The diagram shows part of the curve with equation y2=x2y^{2} = x - 2 and the lines x=5x = 5 and y=1y = 1. The shaded region is enclosed by the curve and the lines is rotated through 360360^{\circ} about the xx-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=1y = 1. For that reason, we also need to consider the cylinder formed by the line y=1y = 1 when it rotates,
V=πy2 dxArea of cylinderV = \int \pi y^{2}\ dx - \textmd{Area of cylinder}Let’s first find the volume produced by the curve,
πy2 dx\int \pi \textcolor{#2192ff}{y^{2}}\ dxSubstitute in y2y^{2},
π(x2) dx\int \pi \textcolor{#2192ff}{(x - 2)}\ dxFrom the diagram, we can tell that one of our limits is 55. The other limit is at y=1y = 1, so we need to convert it to be in terms of xx,
At y=1\textmd{At } y = 1y2=x2y^{2} = x - 212=x21^{2} = x - 21=x21 = x - 2x=3x = 3Therefore, our limits are 33 and 55,
35π(x2) dx\int_{3}^{5} \pi (x - 2)\ dxNow let’s integrate,
π35(x2) dx\pi \int_{3}^{5} (x - 2)\ dxπ[x222x]35\pi \left[\frac{x^{2}}{2} - 2x\right]_{3}^{5}Substitute in the limits,
π[((5)222(5))((3)222(3))]\pi \left[\left(\frac{(5)^{2}}{2} - 2(5)\right) - \left(\frac{(3)^{2}}{2} - 2(3)\right)\right]π[52(32)]\pi \left[\frac{5}{2} - \left(-\frac{3}{2}\right)\right]π[52+32]\pi \left[\frac{5}{2} + \frac{3}{2}\right]4π4\piNow let’s find the volume produced by the cylinder due to the line y=1y = 1,
V=πr2 hV = \pi r^{2}\ hV=π×12×(53)V = \pi \times 1^{2} \times (5 - 3) V=2πV = 2\piTherefore,
V=4π2πV = 4\pi - 2\piV=2πV = 2\piTherefore, the final answer is,
V=2πV = 2\pi