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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.

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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

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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

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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

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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

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(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

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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

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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).

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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

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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)

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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

−

+ − = + − − − +

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= 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