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Imagine a solid object situated as in Fig 82 Observe the axes in the diagram, and imagine that we slice the gure with slices that are vertical (ie, that rise out of the x-y plane) and that are perpendicular to the x-axis (and parallel to the y-axis) Look at Fig 83 Notice, in the gure, that the gure extends from x = a to x = b
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y = f (x)
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Fig 81
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If we can express the area of the slice at position x as a function A(x) of x, then (see Fig 84) the volume of a slice of thickness x at position x will be about A(x) x If P = {x0 , x1 , , xk } is a partition of the interval [a, b] then the volume of the original solid object will be about V =
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Fig 82
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A(xj )
CHAPTER 8 Applications of the Integral
Fig 84
As the mesh of the partition becomes ner and ner, this (Riemann) sum will tend to the integral
A(x) dx
We declare the value of this integral to be the volume V of the solid object
EXAMPLES
EXAMPLE 81
Calculate the volume of the right circular cone with base a disc of radius 3 and height 6
SOLUTION Examine Fig 85 We have laid the cone on its side, so that it extends from x = 0 to x = 6 The upper edge of the gure is the line y = 3 x/2 At position x,
Fig 85
Applications of the Integral
the height of the upper edge is 3 x/2, and that number is also the radius of the circular slice at position x (Fig 86) Thus the area of that slice is A(x) = 3 x 2
thickness , x 3 _ x/2
Fig 86
We nd then that the volume we seek is V =
A(x) dx =
3
dx =
2(3 x/2)3 3
= 18
You Try It: Any book of tables (see [CRC]) will tell you that the volume of a right circular cone of base radius r and height h is 1 r 2 h This formula is consistent 3 with the result that we obtained in the last example for r = 3 and h = 6 Use the technique of Example 81 to verify this more general formula EXAMPLE 82
A solid has base the unit disk in the x-y plane The vertical cross section at position x is an equilateral triangle Calculate the volume
SOLUTION Examine Fig 87 The unit circle has equation x 2 +y 2 = 1 For our purposes, this is more conveniently written as y = 1 x 2 ( )
Thus the endpoints of the base of the equilateral triangle at position x are the points (x, 1 x 2 ) In other words, the base of this triangle is b = 2 1 x 2 Examine Fig 88 We see that an equilateral triangle of side b has height 3b/2 Thus the area of the triangle is 3b2 /4 In our case then, the equilateral
CHAPTER 8 Applications of the Integral
Fig 87
3 b 2
Fig 88
triangular slice at position x has area 2 3 2 1 x 2 = 3(1 x 2 ) A(x) = 4 Finally, we may conclude that the volume we seek is V = =
1 1 1 1
A(x) dx 3(1 x 2 ) dx
1 1
x3 = 3 x 3 1 = 3 1 3 4 3 = 3
( 1)
1 3
EXAMPLE 83
Applications of the Integral
A solid has base in the x-y plane consisting of a unit square with center at the origin and vertices on the axes The vertical cross-section at position x is itself a square Refer to Fig 89 What is the volume of this solid
Fig 89
SOLUTION It is suf cient to calculate the volume of the right half of this solid, and to double the answer Of course the extent of x is then 0 1/ 2 At position x x, the height of the upper edge the square base is 1/ 2 x So the base of of the vertical square slice is 2(1/ 2 x) (Fig 810) The area of the slice is then 2 2 = 2 2x A(x) = 2 1/ 2 x
2 (1/ 2 _ x)
2 (1/ 2 _ x)
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