1. Volumes
© James Taylor 2000
Dove Presentations

2. y
y = g(x)
y = f(x) y
Q(x,y) x O a b
© James Taylor 2000

3. y dx
P(x,y)
y = f(x) x a b O

4. y
y = f(x) b
P(x,y) dy a x O

5. y dx
y = g(x)
P(X,Y)
y = f(x)
Q(x,y) x O a b

6. A good revision problem is to consider the remaining volume after a hole of radius a units is drilled through a sphere of radius r units, with the axis of the hole being one of the diameters of the sphere.
We first need to calculate the abscissae of Q and Q’.

7. When y = a, x2 + a2 = 4a2, so
Now consider a slice through the figure at a point P(x, y) on the circle.

10. The area bounded by the curve y = x2 and the lines x = 0, y = 1 is rotated about the line y = 1. Find the volume of the solid of revolution. y
y = 1 x O

11. Take a vertical slice at a point P(x, y) on the curve
Take a vertical slice at a point P(x, y) on the curve. Its radius at P is R = 1 - y. y
y = 1
P(x, y) x O

12. y
© James Taylor 2000

20. y

22. A slightly different method

28. y

30. An
“aide memoire”

36. © James Taylor 2000

37. We wish to find the volume of a solid whose base is an ellipse, and whose cross-sections perpendicular to the base and the major axis are squares.

38. Taking the equation of the ellipse as
construct axes as shown. y x O

39. Consider a point P (x, y) on the ellipse.

40. Construct an element of volume at P (x, y).

41. Since the indicated length is y units
P(x, y) y O x

42. each edge of the square has length 2y units.
P(x, y) y O x

43. 2y 2y
P(x, y) y O x

44. Hence the area of the square face is 4y2 units2.
Area = 4y2 2y
P(x, y) y O x

45. Dx DV = 4y2Dx O x P(x, y) If the thickness of the element is Dx,
then the volume DV of the prism is
DV = 4y2Dx
P(x, y) O x

46. The more elements of volume we consider, the more accurate our approximation will be.

47. x

48. x

49. A = 4y2 DV = 4y2Dx Area of a square face at P(x, y):
Volume of the square prism at P:
DV = 4y2Dx -a a

50. x
The prisms will be bounded by the curves indicated.

51. The finished solid.

52. Hence the required volume is determined by the integral:

53. Since x a -a

54. We shall now consider some more complex cross-sections.
In the next solid, cross-sections perpendicular to the x axis of a horn shaped solid are circles with diameters running from the curve
to the curve
The solid extends from the origin to its cross-section at x = 1.
We wish to calculate the volume so formed.
You might like to try to draw this solid before advancing to the next slide.

55. y x

56. The radius of the slice is thenx y
The radius of the slice is then
The area of the slice is then

57. We shall now repeat the exercise, but take the cross-sections as equilateral triangles.
It is useful to know (and remember) that the area of an equilateral triangle in terms of its side lengths s is
and in terms of its height h is
Again, you might like to try to draw this solid before advancing to the next slide.

58. y x

59. The equilateral triangle with height h isy
The equilateral triangle with height h is x
From above we know that
Hence

60. In the next example, the base of the solid is the ellipse
Each section of the solid perpendicular to the base and the x axis is a parabola, whose height equals its base width
We wish to calculate the volume so formed.
You might like to try to draw this solid before advancing to the next slide.

61. y x
Base is the ellipse

62. Hence the element of volume at P is:
At the point P(x,y) on the ellipse, the parabola will have a base length of 2y units - and this will also be its height. The exact area of the parabola is given by Simpson’s Rule:
Hence the element of volume at P is: x y

63. Consider now two intersecting pipes.

64. Rotate the cylinders and try to imagine their common volume
Click centre of solid to start, click solid to stop, click here
to advance

65. This is the common volume we are trying to find.
Click centre of solid to start, click solid to stop, click here
to advance.

67. Slices perpendicular to the horizontal are squares.

69. The common volume of the two pipes is related to our first problem.
We can determine our new volume from that result, but it is instructive to work the process again..

70. Take axes as shown, and consider a point P on the circumference of the circle at the red square.y
P(x,y) x
Let the circle be

71. Then the area of the red square is again 4y2.
P(x,y) x

72. If this element of volume has thickness Dx, then
(a result known to Archimedes).

74. © James Taylor 2000

75. In the following typical exercise, no axes or pronumerals are given
In the following typical exercise, no axes or pronumerals are given. All axes must be provided by the reader.
It is not uncommon for similar triangles to be used in such problems.
A tank is constructed so that all cross-sections perpendicular to the vertical axis are hexagons, with sides varying uniformly from 6m to 10m.
If the tank is 12m high, find its volume.
You might like to try to draw this solid before advancing to the next slide.

76. At height h above the base, take a slice of thickness dh

77. A hexagon is made up of 6 equilateral triangles.
This means that if an edge is x m, then so is R.

78. Add the given numerical data:6 4 R h 6

79. We need a relationship between R and h.6 4 4 R
R-6 12 h h 6

80. Use the two similar triangles.6 4 4 12 R
R-6 h h 6

81. The area of cross-section A is 6 equilateral triangles of side width R:4
and we have
R-6 so 12 h

82. A Further Example dt y x
Q(x, y) s t
© James Taylor 2000

83. A dt dt The 1984 HSC 4U paper had a question which repays close study.
The curve y = f(x) is even, and the area enclosed by the curve and the and the x axis is A. y x O s
This area is rotated about the line x = - s. t t A dt dt
The volume formed is to be found by slicing A into thin vertical strips, rotating these to obtain cylindrical shells, and adding the shells.
Two typical strips of width dt whose centre lines are distance t from the y axis are shown.

84. This is not easy to visualise, so the diagram below may help.

85. A dt Prove that the inner shell has approximate volumey x O A s t
Note that the strip as drawn in the question is not rectangular. We are told that the distance t is measured to the centre line of the strip, so we need to point out that the area of the strip is approximately equal to the rectangle drawn at the midpoint as shown above.

86. Let P(x, y) be the point on the curve which is the midpoint of the top edge of the rectangle. Let its distance from the line x = -s be r. y
Then the circumference of the circle through P is s
P(x, y)
Hence the area of the base is t r
The height of the cylinder is x dt O dt
Hence the volume of the cylindrical shell drawn above is
Note: as proved before, the formula for A is exact if the circumference is taken at the mid-point. The approximation comes in here because the original strip was not a rectangle.

87. Q(x, y) dt Prove that the outer shell has approximate volume
At Q(x, y) (the midpoint of the edge) the radius of the cylinder is R = s + t. y t s
Q(x, y) x dt
As before then, the area of the base is
The height of the cylindrical shell is
Hence the volume of the cylindrical shell is

88. Prove that the volume of the solid is
The sum of the two shells found so far is
which simplifies to give
since, being an even function f(t) = f(-t). We only need x to run from 0 to a, so
But
Hence

89. s A

90. This result is a special case of the general result of Pappus, that if an area is moved along a path then the volume of the solid formed is found by multiplying the area of the shape, and the length of the path travelled by the centroid of the shape.
Applying this result to the torus, we find that
Some interesting consequences follow from Pappus’ result. For instance:

91. C
This solid
has the same volume as this solid C
if C is the centroid.

92. This completes the presentation
This completes the presentation. If you enjoyed the presentation or would like to make a comment, send me a note:
Peace in every step