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© 2014, John Bird
1154
CHAPTER 74 VOLUMES OF SOLIDS OF REVOLUTION
EXERCISE 288 Page 783
1. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis: y = 5x ; x = 1, x = 4 A sketch of y = 5x is shown below.
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = ( )4 4 422 2
1 1 1d 5 d 25 dy x x x x xπ π π= =∫ ∫ ∫
= 43
1
64 1 6325 25 253 3 3 3xπ π π = − =
= 525π cubic units
2. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis: y = x2 ; x = –2, x = 3 A sketch of y = 2x is shown below.
© 2014, John Bird
1155
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = ( )3 3 322 2 42 2 2
d d dy x x x x xπ π π− − −
= =∫ ∫ ∫
= 35
2
243 32 2755 5 5 5xπ π π
−
= − − = = 55π cubic units
3. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis:
y = 2x3 + 3 ; x = 0, x = 2
A sketch of y = 32 3x + is shown below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = ( ) ( )2 2 222 2 4 2
0 0 0d 2 3 d 4 12 9 dy x x x x x xπ π π= + = + +∫ ∫ ∫
= ( )25 3
0
4 12 1289 32 18 05 3 5x x xπ π + + = + + −
= ( )25.6 32 18π + + = 75.6π cubic units
4. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis:
2
4y = x ; x = 1, x = 5
A sketch of 2
4y x= , i.e. 2 4y x= is shown below
© 2014, John Bird
1156
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = 525 5
21 1
1
25 1d 4 d 4 42 2 2xy x x xπ π π π = = = −
∫ ∫
= 48π cubic units
5. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis: xy = 3 ; x = 2, x = 3
A sketch of xy = 3, i.e. 3yx
= is shown below.
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = 2
3 3 3 32 2
22 2 2 2
3 9d d d 9 dy x x x x xx x
π π π π − = = = ∫ ∫ ∫ ∫
= 3 31
2 2
1 1 1 19 9 9 91 3 2 6
xx
π π π π− = − = − − = − − −
= 1.5π cubic units
© 2014, John Bird
1157
6. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curves, the y-axis and the given ordinates through one revolution about the y-axis: y = x2 ; y = 1, y = 3
A sketch of y = x2 is shown below
When the shaded area is rotated one revolution about the y-axis
volume = 3
21
dx yπ∫
Since y = x2 , then x = y
Hence, volume = ( )323 3
21 1
1
( ) d d (4.5) 0.52yy y y xπ π π π = = = − ∫ ∫ = 4π cubic units
7. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curves, the y-axis and the given ordinates through one revolution about the y-axis: y = 3x2 – 1; y = 2, y = 4 A sketch of y = 23 1x − , is shown below
© 2014, John Bird
1158
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution
given by: volume = ( ) ( )424 4
22 2
2
1d d 8 4 2 23 3 2 3
y yx y y yπ ππ π + = = + = + − + ∫ ∫
= [ ]83π = 2.67π cubic units
8. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curves, the y-axis and the given ordinates through one revolution about the y-axis:
y = 2x
; y = 1, y = 3
A sketch of y = 2x
, is shown below.
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution
given by: volume = ( )2 313 3 3
2 21 1 1
1
2 1d d 4 d 4 4 11 3
yx y x y yy
π π π π π−
− = = = = − − −
∫ ∫ ∫
= 243
π − − = 2.67π cubic units
9. The curve y = 2x2 + 3 is rotated about (a) the x-axis between the limits x = 0 and x = 3, and (b) the y-axis, between the same limits. Determine the volume generated in each case. (a) The curve is shown below.
( ) ( )353 322 4 2 3-
0 00
4Volume 2 3 d 4 12 9 d 4 95
x axisxx x x x x x xπ π π = + = + + = + + ∫ ∫
= π[(329.4) – (0)] = 329.4π cubic units
© 2014, John Bird
1159
(b)
212-
3Volume dy axis x yπ= ∫ Since y = 2x2 + 3, then 2x2 = y – 3 and 2 3
2yx −
=
Hence, 212 221
-3
3
3 21 9volume d 3 63 92 2 2 2 2 2
y axisy yy yπ ππ − = = − = − − −
∫
= ( ) ( ) 162157.5 4.52 2π π − − =
= 81π cubic units
© 2014, John Bird
1160
EXERCISE 289 Page 785
1. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis: y = 4ex ; x = 0, x = 2
A graph of y = 4ex lies wholly above the x-axis. Revolving y = 4ex one revolution about the x-axis produces a solid of revolution given by:
volume = ( ) [ ]222 2 222 2 4 0
0 0 00
ed 4e d 16 e d 16 8 e e2
xx xy x x xπ π π π π = = = = − ∫ ∫ ∫
= 428.8π cubic units
2. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curve, the x-axis and the given ordinates through one revolution about the x-axis:
y = sec x ; x = 0, x = 4π
A sketch of part of y = sec x is shown below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = [ ]/4 /4 /42 2
00 0d sec d tan tan tan 0
4y x x x x
π π π ππ π π π = = = − ∫ ∫
= [ ]1 0π − = π cubic units
3. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curves, the y-axis and the given ordinates through one revolution about the y-axis: x2 + y2 = 16 ; y = 0, y = 4
© 2014, John Bird
1161
A sketch of x2 + y2 = 16 , i.e. x2 + y2 = 42 is shown below
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution
given by: volume = ( ) ( )434 4
2 20 0
0
64d 16 d 16 64 03 3yx y y y yπ π π π = − = − = − −
∫ ∫
= 42.67π cubic units
4. Determine the volume of the solid of revolution formed by revolving the area enclosed by the curves, the y-axis and the given ordinates through one revolution about the y-axis: x y = 2 ; y = 2, y = 3
A sketch x y = 2, i.e. 2yx
= and y = 2
4x
is shown below.
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution
given by: volume = [ ] [ ]3 3 32
22 2
4 3d d 4 ln 4 ln 3 ln 2 4 ln2
x y y yy
π π π π π= = = − =∫ ∫
= 1.622π cubic units
5. Determine the volume of a plug formed by the frustum of a sphere of radius 6 cm which lies between two parallel planes at 2 cm and 4 cm from the centre and on the same side of it (the equation of a circle, centre 0, radius r is x2 + y2 = r2).
© 2014, John Bird
1162
The volume of a frustum of a sphere is determined by rotating the curve 2 2 26x y+ = one revolution
about the x-axis, between the limits x = 2 and x = 4, i.e. rotating the shaded area of the diagram
below
Volume of frustum = ( )434 4
2 2 22 2
2
64 2d 6 36 144 723 3 3xy x x dx xπ π π π = − = − = − − −
∫ ∫
= 53.33π cubic units
6. The area enclosed between the two curves x2 = 3y and y2 = 3x is rotated about the x-axis. Determine the volume of the solid formed. The curves are shown in the sketch below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = ( )34 2 53 3
20 0
0
3 27 243d 3 d 09 2 45 2 45x x xy x x xπ π π π = − = − = − −
∫ ∫
= [ ]13.5 5.4π − = 8.1π cubic units
© 2014, John Bird
1163
7. The portion of the curve y = x2 + 1x
lying between x = 1 and x = 3 is revolved 360° about the x-
axis. Determine the volume of the solid formed.
A sketch of part of 2 1y xx
= + , is shown below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by: volume = 2 35 13 3 3
2 2 4 221 1 1
1
1 1d d 2 d5 1x xy x x x x x x x
x xπ π π π
− = + = + + = + + − ∫ ∫ ∫
= 243 1 19 1 15 3 5
π + − − + − = 57.07π cubic units
8. Calculate the volume of the frustum of a sphere of radius 5 cm that lies between two parallel planes at 3 cm and 2 cm from the centre and on opposite sides of it. The volume of a frustum of a sphere may be determined by integration by rotating the curve
x2 + y2 = 52 (i.e. a circle, centre 0, radius 5) one revolution about the x-axis, between the limits x = 3
and x = –2 (i.e. rotating the shaded area of sketch below)
© 2014, John Bird
1164
Volume of frustum = 333 3
2 2 22 2
2
8d (5 )d 25 (75 9) 503 3xy x x x xπ π π π
− −−
= − = − = − − − + ∫ ∫
= 113.33π cubic units
9. Sketch the curves y = x2 + 2 and y – 12 = 3x from x = –3 to x = 6. Determine (a) the coordinates of the points of intersection of the two curves, and (b) the area enclosed by the two curves. (c) If the enclosed area is rotated 360° about the x-axis, calculate the volume of the solid produced. The curves are shown sketched below
(a) Equating the y-values gives: 2 2x + = 3x + 12
i.e. 2 3 10 0x x− − =
and (x – 5)(x + 2) = 0
from which, x = 5 and x = –2
When x = –2, y = 6 and when x = 5, y = 27
Hence, the points of intersection of the two curves are at (–2, 6) and (5, 27)
(b) Shaded area = ( ) ( ) ( )5 5 5
2 22 2 2
d 3 12 2 d 3 10 dy x x x x x x x− − −
= + − + = + − ∫ ∫ ∫
= 52 3
2
3 75 125 810 50 6 202 3 2 3 3x xx
−
+ − = + − − − +
© 2014, John Bird
1165
= 57.17 square units
(c) Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution
given by:
volume =
( ) ( ) ( ) ( )5 5 52 22 2 2 4 22 2 2
d 3 12 2 d 9 72 144 4 4 dy x x x x x x x x xπ π π− − −
= + − + = + + − + + ∫ ∫ ∫
= ( )53 2 55
2 42
2
5 725 72 140 d 1403 2 5x x xx x x x xπ π
−−
+ + − = + + − ∫
= 625 40 3236(25) 140(5) 625 144 2803 3 5
π + + − − − + − +
= ( ) ( )208.33 900 700 625 13.33 144 280 6.4π + + − − − + − +
= 1326π cubic units