15.2. Volumes of revolution. Introduction. Prerequisites. Learning Outcomes. Learning Style

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1 Volumes of revolution 15.2 Introduction In this block we show how the concept of integration as the limit of a sum can be used to find volumes of solids formed when curves are rotated around the x or y axes. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... calculate volumes of revolution 1 be able to calculate definite integrals 2 understand integration as the limit of a sum Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises

2 1. Volumes generated by rotating curves about the x-axis Figure 1 shows a graph of the function y =2x for x between 0 and 3. 6 y y =2x O 3 x Figure 1. A graph of the function y =2x, for 0 x 3. Imagine rotating the line y =2x by one complete revolution (360 0 or 2π radians) around the x-axis. The surface so formed is the surface of a cone as shown in Figure 2. Such a threedimensional shape is known as a solid of revolution. We now discuss how to obtain the volumes of such solids of revolution. 6 y y =2x O 3 x Figure 2. When the line y =2x is rotated around the axis, a solid is generated. Try each part of this exercise Find the volume of the cone generated by rotating y =2x, for 0 x 3, around the x-axis, as shown in Figure 2. In order to find the volume of this solid we assume that it is composed of lots of thin circular discs all aligned perpendicular to the x-axis, such as that shown in Figure 3. From Figure 3 we Engineering Mathematics: Open Learning Unit Level 1 2

3 note that a typical disc has radius y, which in this example equals 2x, and thickness δx. y 6 y =2x (x, y) δx O 3 x Figure 3. The cone is divided into a number of thin circular discs. The volume of a circular disc is the circular area multiplied by the thickness. Part (a) Write down an expression for the volume of this typical disc: To find the total volume we must sum the contributions from all discs and find the limit of this sum as the number of discs becomes infinite and δx becomes zero. That is This is the definition of a definite integral. x=3 lim δx 0 x=0 4πx 2 δx Part (b) Write down the corresponding integral. Part (c) Find the required volume by performing the integration: Now let us do another example. Try each part of this exercise A graph of the function y = x 2 for x between 0 and 4 is shown in Figure 4. The graph is rotated around the x-axis to produce the solid shown. Find its volume. 3 Engineering Mathematics: Open Learning Unit Level 1

4 y 16 y = x 2 (x, y) O δx 4 x Figure 4. The solid of revolution is divided into a number of thin circular discs. As in the previous guided exercise, the solid is considered to be composed of lots of circular discs of radius y, (which in this example is equal to x 2 ), and thickness δx. Part (a) Write down the volume of each disc: Part (b) Write down the expression which results by summing the volumes of all such discs: Part (c) Write down the integral which results from taking the limit of the sum as δx 0: Part (d) Perform the integration to find the volume of the solid: Try each part of this exercise In general, suppose the graph of y(x) between x = a and x = b is rotated about the x-axis, and the solid so formed is considered to be composed of lots of circular discs of thickness δx. Part (a) Write down an expression for the radius of a typical disc: Part (b) Write down an expression for the volume of a typical disc: The total volume is found by summing these individual volumes and taking the limit as δx tends to zero: πy 2 δx x=b lim δx 0 x=a Engineering Mathematics: Open Learning Unit Level 1 4

5 Part (c) Write down the definite integral which this sum defines: Key Point If the graph of y(x), between x = a and x = b, is rotated about the x-axis the volume of the solid formed is b πy 2 dx a More exercises for you to try 1. When the graph of y(x) between x = a and x = b is rotated around the x-axis, show that the volume of the solid formed is b a πy2 dx. 2. Find the volume of the solid formed when that part of the curve between y = x 2 between x = 1 and x = 2 is rotated about the x-axis. 3. The parabola y 2 =4x for 0 x 1, is rotated around the x-axis. Find the volume of the solid formed. 2. Volumes generated by rotating curves about the y-axis We can obtain a different solid of revolution by rotating a curve around the y-axis instead of around the x-axis. See Figure 5. y y(x) δy (x, y) O x Figure 5. A solid can be generated by rotation around the y-axis. To find the volume of this solid it is divided into a number of circular discs as before, but this time the discs are horizontal. The radius of a typical disc is x and its thickness is δy. The volume of the disc will be πx 2 δy where δy is the thickness of the disc. The total volume is found by summing these individual volumes and taking the limit as δy 0. If the lower and upper limits on y are c and d, we obtain for the volume: y=d lim δy 0 y=c πx 2 δy 5 Engineering Mathematics: Open Learning Unit Level 1

6 which is the definite integral d c πx 2 dy Key Point If the graph of y(x), between y = c and y = d, is rotated about the y-axis the volume of the solid formed is d πx 2 dy c Try each part of this exercise Find the volume generated when the graph of y = x 2 between x = 0 and x = 1 is rotated around the y-axis. On the graph of y = x 2, when x =0,y = 0 and when x =1y = 1 and so the limits on y are the same as the limits on x. Part (a) Write down the required integral. Part (b) Because y = x 2 this integral can be written entirely in terms of y. Do this now, and then evaluate the integral. More exercises for you to try 1. When the graph of y(x) between x = a and x = b is rotated around the y-axis, show that the volume of the solid formed is d c πx2 dy where c = y(a) and d = y(b). 2. The curve y = x 2 for 1 <x<2 is rotated about the y-axis. Find the volume of the solid formed. 3. The line y =2 2x for 0 x 2 is rotated around the y-axis. Find the volume of revolution. Engineering Mathematics: Open Learning Unit Level 1 6

7 3. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. DERIVE can be used to obtain definite integrals. In particular there are specific commands for determining volumes of revolution obtained when curves are rotated about either the x- or y-axes. These commands can be obtained by opening the Int apps library (simply click on File:Open and then double click the Int apps icon). The commands take the form Volume of revolution(y, x, x 1,x 2 ) if y(x) x 1 x x 2 is rotated round the x-axis or, Volumey of revolution(y, x, x 1,x 2 ) if y(x) x 1 x x 2 is rotated round the y-axis For example to find the volume of revolution obtained by rotating the curve y = x 2 1 x 2 about the x-axis we would first open Library Int apps and then key Author:Expression, then type: Volume of revolution(x 2,x,1, 2). DERIVE responds VOLUME OF REVOLUTION(x 2,x,1, 2) Then hit Simplify:Basic and DERIVE responds 31 π 5 However, beware, the volume of revolution about the y-axis as provided by DERIVE is not the same as that given in this Block. In fact, referring to the following diagram, DERIVE gives the outer volume of revolution whereas, in the text, we calculate the inner volume of revolution. y y x x outer volume inner volume outer volume x2 x 1 2πxydx inner volume x2 x 1 πx 2 dy Of course, the two measures are related: outer volume + inner volume = π(x 2 2y 2 x 2 1y 1 ) 7 Engineering Mathematics: Open Learning Unit Level 1

8 End of Block15.2 Engineering Mathematics: Open Learning Unit Level 1 8

9 π(2x) 2 δx =4πx 2 δx 9 Engineering Mathematics: Open Learning Unit Level 1

10 3 0 4πx2 dx Engineering Mathematics: Open Learning Unit Level 1 10

11 [ ] 3 4πx 3 =36π Engineering Mathematics: Open Learning Unit Level 1

12 π(x 2 ) 2 δx = πx 4 δx Engineering Mathematics: Open Learning Unit Level 1 12

13 x=4 x=0 πx4 δx 13 Engineering Mathematics: Open Learning Unit Level 1

14 4 0 πx4 dx Engineering Mathematics: Open Learning Unit Level 1 14

15 4 5 π 5 = 204.8π 15 Engineering Mathematics: Open Learning Unit Level 1

16 y Engineering Mathematics: Open Learning Unit Level 1 16

17 πy 2 δx 17 Engineering Mathematics: Open Learning Unit Level 1

18 b a πy2 dx Engineering Mathematics: Open Learning Unit Level 1 18

19 2. 31π/5, 3. 2π. 19 Engineering Mathematics: Open Learning Unit Level 1

20 1 0 πx2 dy Engineering Mathematics: Open Learning Unit Level 1 20

21 [ ] 1 1 πydy = πy 2 = π Engineering Mathematics: Open Learning Unit Level 1

22 2. 15π 16π Engineering Mathematics: Open Learning Unit Level 1 22

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