CHAPTER 74 VOLUMES OF SOLIDS OF REVOLUTION
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 == ∫ π y 2 d x π= ∫ ( 5 x ) d x π ∫ 25 x 2 d x 4
4
1
4
2
1
1
64 1 x3 63 = 25π = 25 π −= 25π = 525π cubic units 3 1 3 3 3 4
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 = x 2 is shown below.
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© 2014, John Bird
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume == ∫ π y 2 d x π= ∫ ( x2 ) d x π ∫ x4 d x 3
3
−2
−2
3
2
−2
243 32 x5 275 = π = − π π −= = 55π cubic units 5 −2 5 5 5 3
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 = 2 x3 + 3 is shown below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume =
∫
2 0
π y 2 d= x π ∫ ( 2 x 2 + 3) d= x π ∫ ( 4 x 4 + 12 x 2 + 9 ) d x 2
2
2
0
0
128 4 x5 12 x3 =π x π + + 9= + 32 + 18 − ( 0 ) 3 5 0 5 2
= π ( 25.6 + 32 + 18 ) = 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: y2 =x; 4
A sketch of
x = 1, x = 5
y2 = x , i.e. y 2 = 4 x is shown below 4
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© 2014, John Bird
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume =
5 25 1 x2 2dx = = = y 4 x d x 4 4π − π π π ∫1 ∫1 2 2 2 1 5
5
= 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. y =
3 is shown below. x
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
2
3 9 3 3 volume = ∫= d x π= d x 9 π y dx π∫= π ∫2 x2 ∫ 2 x −2 d x 2 2 x 3
2
3
3
3
x −1 1 1 1 1 = 9π = −9π = −9π − = −9π − −1 2 x2 3 2 6 = 1.5π cubic units 1156
© 2014, John Bird
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
1
π x2 d y
Since y = x2 , then x =
y 3
y2 Hence, volume = ∫ π ( = y ) d y π ∫= y d x π= 2 π (4.5) − ( 0.5 ) = 4π cubic units 1 1 1 3
2
3
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 = 3 x 2 − 1 , is shown below
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© 2014, John Bird
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution given by:
π y2 π y +1 ( 8 + 4 ) − ( 2 + 2 ) volume = ∫ π x d = y π∫ d y = + y = 2 2 32 3 2 3 4
4
4
2
=
π 3
[8] = 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=
2 ; x
y = 1, y = 3
A sketch of y =
2 , is shown below. x
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution 2
given by:
3 2 1 y −1 volume = ∫ π x d y = = −4π − (1) π∫ dx = 4π ∫ y −2 d y = 4π 1 1 1 −1 1 3 y 3
3
3
2
2 = −4π − = 2.67π cubic units 3
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.
= Volume x -axis
4 x5
3
∫0 π ( 2 x 2 + 3) d=x π ∫0 ( 4 x 4 + 12 x 2 + 9 ) d=x π 5 + 4 x3 + 9 x 0 3
2
3
= π[(329.4) – (0)] = 329.4π cubic units 1158
© 2014, John Bird
21
(b) Volume y -axis = ∫ π x 2 d y Hence, volume y -axis = π∫
2x2 = y – 3 and x 2 =
Since y = 2x2 + 3, then
3
21 3
y −3 2
π y2 π 212 y −3 9 d = y − 3 y = − 63 − − 9 22 2 2 2 3 2 21
=
π
162π (157.5 ) − ( −4.5 ) = 2 2
= 81π cubic units
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© 2014, John Bird
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: 2
e2 x volume = ∫ π y= d x π ∫ ( 4 e )= d x 16π ∫ e = d x 16π = 8π [ e 4 − e0 ] 0 0 0 2 0 2
2
2
x
2
2
2x
= 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 =
π 4
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 =
∫
π /4 0
π= y2 d x π ∫
π /4 0
π /4 π 2 xd x = = π [ tan π tan − tan 0 sec x] 0 4
= π [1 − 0] = 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 1160
© 2014, John Bird
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:
y3 64 volume = ∫ π x d y= π ∫ (16 − y ) d y= π 16 y − = π 64 − − ( 0 ) 0 0 3 0 3 4
4
2
4
2
= 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
y=
A sketch x y = 2, i.e.
2 4 and y = is shown below. x2 x
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution given by:
volume =
∫
3 2
π x 2 d y= π ∫
3 2
4 3 3 d y= 4π [ ln y ] = 4π [ ln 3 − ln 2= ] 4π ln 2 y 2 = 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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© 2014, John Bird
The volume of a frustum of a sphere is determined by rotating the curve x 2 + y 2 = 62 one revolution about the x-axis, between the limits x = 2 and x = 4, i.e. rotating the shaded area of the diagram below
x3 64 2 Volume of frustum = ∫ π y d = = π 36 x − = π 144 − − 72 − x π ∫ ( 6 − x ) dx 2 2 3 2 3 3 4
2
4
4
2
2
= 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 =
∫
3 0
0 3
π y 2 d x = π ∫ 3x −
27 243 x4 3x 2 x5 − = π − d x= π − ( 0 ) 9 45 2 45 0 2 3
= π [13.5 − 5.4] = 8.1π cubic units
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© 2014, John Bird
7. The portion of the curve y = x2 +
1 lying between x = 1 and x = 3 is revolved 360° about the xx
axis. Determine the volume of the solid formed.
A sketch of part of = y x2 +
1 , is shown below x
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
2
3
3 1 1 x −1 x5 volume = ∫ π y d x = π ∫ x 2 + d x = π ∫ x 4 + 2 x + d x = π + x 2 + 1 1 1 −1 1 x x2 5 3
2
3
243 1 1 = π + 9 − − + 1 − 1 = 57.07π cubic units 3 5 5
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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© 2014, John Bird
x3 8 Volume of frustum = ∫ π y = d x ∫ π (5 − x )= d x π 25 x − = π (75 − 9) − −50 + −2 −2 3 −2 3 3
3
2
3
2
2
= 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:
x 2 + 2 = 3x + 12
x 2 − 3 x − 10 = 0
i.e. 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 −2
y d= x
∫ −2 ( 3x + 12 ) − ( x 2 + 2 ) d=x 5
∫ ( 3x + 10 − x ) d x 5
2
−2
75 x3 125 8 3x 2 = + 10 x − = + 50 − − 6 − 20 + 3 − 2 2 3 3 2 5
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© 2014, John Bird
= 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 −2
π y2 = dx π∫
( 3x + 12 ) − ( x 2 + 2 ) −2 5
2
2
d x π ∫ ( 9 x 2 + 72 x + 144 ) − ( x 4 + 4 x 2 + 4 ) d x = −2 5
5
x5 5 x3 72 x 2 = π ∫ ( 5 x + 72 x + 140 − x ) d = + + 140 x − x π −2 2 5 −2 3 5
2
4
625 32 40 = π + 36(25) + 140(5) − 625 − − + 144 − 280 + 5 3 3 = π ( 208.33 + 900 + 700 − 625 ) − ( −13.33 + 144 − 280 + 6.4 ) = 1326π cubic units
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© 2014, John Bird
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 == ∫ π y 2 d x π= ∫ ( 5 x ) d x π ∫ 25 x 2 d x 4
4
1
4
2
1
1
64 1 x3 63 = 25π = 25 π −= 25π = 525π cubic units 3 1 3 3 3 4
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 = x 2 is shown below.
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© 2014, John Bird
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume == ∫ π y 2 d x π= ∫ ( x2 ) d x π ∫ x4 d x 3
3
−2
−2
3
2
−2
243 32 x5 275 = π = − π π −= = 55π cubic units 5 −2 5 5 5 3
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 = 2 x3 + 3 is shown below
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume =
∫
2 0
π y 2 d= x π ∫ ( 2 x 2 + 3) d= x π ∫ ( 4 x 4 + 12 x 2 + 9 ) d x 2
2
2
0
0
128 4 x5 12 x3 =π x π + + 9= + 32 + 18 − ( 0 ) 3 5 0 5 2
= π ( 25.6 + 32 + 18 ) = 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: y2 =x; 4
A sketch of
x = 1, x = 5
y2 = x , i.e. y 2 = 4 x is shown below 4
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© 2014, John Bird
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
volume =
5 25 1 x2 2dx = = = y 4 x d x 4 4π − π π π ∫1 ∫1 2 2 2 1 5
5
= 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. y =
3 is shown below. x
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
2
3 9 3 3 volume = ∫= d x π= d x 9 π y dx π∫= π ∫2 x2 ∫ 2 x −2 d x 2 2 x 3
2
3
3
3
x −1 1 1 1 1 = 9π = −9π = −9π − = −9π − −1 2 x2 3 2 6 = 1.5π cubic units 1156
© 2014, John Bird
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
1
π x2 d y
Since y = x2 , then x =
y 3
y2 Hence, volume = ∫ π ( = y ) d y π ∫= y d x π= 2 π (4.5) − ( 0.5 ) = 4π cubic units 1 1 1 3
2
3
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 = 3 x 2 − 1 , is shown below
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© 2014, John Bird
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution given by:
π y2 π y +1 ( 8 + 4 ) − ( 2 + 2 ) volume = ∫ π x d = y π∫ d y = + y = 2 2 32 3 2 3 4
4
4
2
=
π 3
[8] = 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=
2 ; x
y = 1, y = 3
A sketch of y =
2 , is shown below. x
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution 2
given by:
3 2 1 y −1 volume = ∫ π x d y = = −4π − (1) π∫ dx = 4π ∫ y −2 d y = 4π 1 1 1 −1 1 3 y 3
3
3
2
2 = −4π − = 2.67π cubic units 3
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.
= Volume x -axis
4 x5
3
∫0 π ( 2 x 2 + 3) d=x π ∫0 ( 4 x 4 + 12 x 2 + 9 ) d=x π 5 + 4 x3 + 9 x 0 3
2
3
= π[(329.4) – (0)] = 329.4π cubic units 1158
© 2014, John Bird
21
(b) Volume y -axis = ∫ π x 2 d y Hence, volume y -axis = π∫
2x2 = y – 3 and x 2 =
Since y = 2x2 + 3, then
3
21 3
y −3 2
π y2 π 212 y −3 9 d = y − 3 y = − 63 − − 9 22 2 2 2 3 2 21
=
π
162π (157.5 ) − ( −4.5 ) = 2 2
= 81π cubic units
1159
© 2014, John Bird
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: 2
e2 x volume = ∫ π y= d x π ∫ ( 4 e )= d x 16π ∫ e = d x 16π = 8π [ e 4 − e0 ] 0 0 0 2 0 2
2
2
x
2
2
2x
= 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 =
π 4
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 =
∫
π /4 0
π= y2 d x π ∫
π /4 0
π /4 π 2 xd x = = π [ tan π tan − tan 0 sec x] 0 4
= π [1 − 0] = 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 1160
© 2014, John Bird
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:
y3 64 volume = ∫ π x d y= π ∫ (16 − y ) d y= π 16 y − = π 64 − − ( 0 ) 0 0 3 0 3 4
4
2
4
2
= 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
y=
A sketch x y = 2, i.e.
2 4 and y = is shown below. x2 x
Revolving the shaded area shown one revolution about the y-axis produces a solid of revolution given by:
volume =
∫
3 2
π x 2 d y= π ∫
3 2
4 3 3 d y= 4π [ ln y ] = 4π [ ln 3 − ln 2= ] 4π ln 2 y 2 = 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).
1161
© 2014, John Bird
The volume of a frustum of a sphere is determined by rotating the curve x 2 + y 2 = 62 one revolution about the x-axis, between the limits x = 2 and x = 4, i.e. rotating the shaded area of the diagram below
x3 64 2 Volume of frustum = ∫ π y d = = π 36 x − = π 144 − − 72 − x π ∫ ( 6 − x ) dx 2 2 3 2 3 3 4
2
4
4
2
2
= 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 =
∫
3 0
0 3
π y 2 d x = π ∫ 3x −
27 243 x4 3x 2 x5 − = π − d x= π − ( 0 ) 9 45 2 45 0 2 3
= π [13.5 − 5.4] = 8.1π cubic units
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© 2014, John Bird
7. The portion of the curve y = x2 +
1 lying between x = 1 and x = 3 is revolved 360° about the xx
axis. Determine the volume of the solid formed.
A sketch of part of = y x2 +
1 , is shown below x
Revolving the shaded area shown one revolution about the x-axis produces a solid of revolution given by:
2
3
3 1 1 x −1 x5 volume = ∫ π y d x = π ∫ x 2 + d x = π ∫ x 4 + 2 x + d x = π + x 2 + 1 1 1 −1 1 x x2 5 3
2
3
243 1 1 = π + 9 − − + 1 − 1 = 57.07π cubic units 3 5 5
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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© 2014, John Bird
x3 8 Volume of frustum = ∫ π y = d x ∫ π (5 − x )= d x π 25 x − = π (75 − 9) − −50 + −2 −2 3 −2 3 3
3
2
3
2
2
= 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:
x 2 + 2 = 3x + 12
x 2 − 3 x − 10 = 0
i.e. 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 −2
y d= x
∫ −2 ( 3x + 12 ) − ( x 2 + 2 ) d=x 5
∫ ( 3x + 10 − x ) d x 5
2
−2
75 x3 125 8 3x 2 = + 10 x − = + 50 − − 6 − 20 + 3 − 2 2 3 3 2 5
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© 2014, John Bird
= 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 −2
π y2 = dx π∫
( 3x + 12 ) − ( x 2 + 2 ) −2 5
2
2
d x π ∫ ( 9 x 2 + 72 x + 144 ) − ( x 4 + 4 x 2 + 4 ) d x = −2 5
5
x5 5 x3 72 x 2 = π ∫ ( 5 x + 72 x + 140 − x ) d = + + 140 x − x π −2 2 5 −2 3 5
2
4
625 32 40 = π + 36(25) + 140(5) − 625 − − + 144 − 280 + 5 3 3 = π ( 208.33 + 900 + 700 − 625 ) − ( −13.33 + 144 − 280 + 6.4 ) = 1326π cubic units
1165
© 2014, John Bird
