ABSTRACT
Problem 2. Find the volume of the solid of revolution when the curve y = 2x is rotated one revolution about the x-axis between the limits x = 0 and x = 5
When y = 2x is revolved one revolution about the x-axis between x = 0 and x = 5 (see Fig. 74.4) then: volume generated
πy2dx = ∫ 5 0
π(2x)2dx
4πx2dx = 4π [ x3
= 500π 3
= 1662 3 π cubic units
of radius 10 and
= 1 3 πr2h = 1
3 π(10)25 = 500π
= 1662 3 π cubic units.)
Problem 3. The curve y = x2 + 4 is rotated one revolution about the x-axis between the limits x = 1 and x = 4. Determine the volume of the solid of revolution produced
Revolving the shaded area shown in Figure 74.5 about the x-axis 360◦ produces a solid of revolution given by:
Volume = ∫ 4 1
πy2 dx = ∫ 4 1
π(x2 + 4)2dx
π(x4 + 8x2 + 16)dx
= π [ x5
5 + 8x
3 + 16x
= π[(204.8+ 170.67+ 64) − (0.2+ 2.67+ 16)]
= 420.6π cubic units
Problem 4. If the curve in Problem 3 is revolved about the y-axis between the same limits, determine the volume of the solid of revolution produced
The volume produced when the curve y = x2 + 4 is rotated about the y-axis between y = 5 (when x = 1)
and y = 20 Figure 74.5 by:
volume = ∫ 20 5
πx2 dy
Since y = x2 + 4, then x2 = y − 4
Hence volume = ∫ 20 5
π(y − 4)dy = π [ y2
2 − 4y
= π[(120)− (−7.5)] = 127.5π cubic units
Now try the following Practice Exercise
Practice Exercise 288 Further problems on volumes of solids of revolution (answers on page 1143)
(Answers are in cubic units and in terms of π .) In Problems 1 to 5, determine the volume of the solid of revolution formed by revolving the areas enclosed by the given curve, the x-axis and the given ordinates through one revolution about the x-axis.
