[PDF] [PDF] Volumes of solids of revolution - Mathcentre

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[PDF] Volumes of solids of revolution - Mathcentre

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Volumes of solids ofrevolution

mc-TY-volumes-2009-1 We sometimes need to calculate the volume of a solid which canbe obtained by rotating a curve about thex-axis. There is a straightforward technique which enables this to be done, using integration. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: •find the volume of a solid of revolution obtained from a simplefunctiony=f(x)between given limitsx=aandx=b; •find the volume of a solid of revolution obtained from a simplefunctiony=f(x)where the limits are obtained from the geometry of the solid.

Contents

1.Introduction 2

2.The volume of a sphere 4

3.The volume of a cone 4

4.Another example 5

5.Rotating a curve about they-axis 6

www.mathcentre.ac.uk 1c?mathcentre 2009

1. IntroductionSuppose we have a curve,y=f(x).

y = f(x) x = a x = b Imagine that the part of the curve between the ordinatesx=aandx=bis rotated about the x-axis through360◦. The curve would then map out the surface of a solid as it rotated. Such solids are calledsolids of revolution. Thus if the curve was a circle, we would obtain the surface of a sphere. If the curve was a straight line through the origin, we would obtain the surface of a cone. Now we already know what the formulae for the volumes of a sphere and a cone are, but where did they come from? How can they calculated? If we could find a general method for calculating the volumes of the solids of revolution thenwe would be able to calculate, for example, the volume of a sphere and the volume of a cone, as well as the volumes of more complex solids. To see how to carry out these calculations we look first at the curve, together with the solid it maps out when rotated through360◦. y = f(x) Now if we take a cross-section of the solid, parallel to they-axis, this cross-section will be a

circle. But rather than take a cross-section, let us take a thin disc of thicknessδx, with the face

of the disc nearest they-axis at a distancexfrom the origin. www.mathcentre.ac.uk 2c?mathcentre 2009 y = f(x) x = a x = b δx xy y + δy

The radius of this circular face will then bey. The radius of the other circular face will bey+δy,

whereδyis the change inycaused by the small positive increase inx,δx. The disc is not a

cylinder, but it is very close to one. It will become even closer to one asδx, and henceδy, tends

to zero. Thus we approximate the disc with a cylinder of thickness, or height,δx, and radiusy. The volumeδVof the disc is then given by the volume of a cylinder,πr2h, so that

δV=πy2δx.

So the volumeVof the solid of revolution is given by

V= limδx→0x=b?

x=aδV = lim

δx→0x=b?

x=aπy

2δx

b a

πy2dx,

where we have changed the limit of a sum into a definite integral, using our definition of inte- gration. This formula now gives us a way to calculate the volumes of solids of revolution about thex-axis.

Key Point

Ifyis given as a function ofx, the volume of the solid obtained by rotating the portion of the curve betweenx=aandx=babout thex-axis is given by V=? b a

πy2dx.

www.mathcentre.ac.uk 3c?mathcentre 2009

2. The volume of a sphereThe equationx2+y2=r2represents the equation of a circle centred on the origin andwith

radiusr. So the graph of the functiony=⎷ r2-x2is a semicircle. r-r y = ⎷r2 - x2quotesdbs_dbs3.pdfusesText_6