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Adding up the force on each R j gives a total force of in .NET framework
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SOLUTION
We let the independent variable h denote depth, measured vertically down from the surface of the water Since the pool is rectangular with vertical sides, w( h) is constantly equal to 20 (because we are interested in the long side) We use 624 pounds per cubic foot for the density of water According to ( ) , the total force on the long side is 8 0 8 0 h 624 w( h) dh =
h 624 20 dh = 39936 lbs
YOU TRY IT A tank full of water is in the shape of a cube of side 10 feet How
much force is exerted against the wall of the tank between the depths of 3 feet and 6 feet
EXAMPLE
A tank has vertical cross section in the shape of an inverted isosceles triangle with horizontal base, as shown in Figure 839 Notice that the base of the tank has length 4 feet and the height is 9 feet The tank is filled with water to a depth of 5 feet Water has density 624 pounds per cubic foot Calculate the total force on one end of the tank SOLUTION
As shown in Figure 840, at depth h (measured down from the surface of the water), the tank has width corresponding to the base of an isosceles triangle similar to the triangle describing the end of the tank The height 4 ft
9 ft 5 ft
FIGURE 839
CALCULUS DeMYSTiFieD
9 5 5_h
FIGURE 840
of this triangle is 5 h Thus we can solve 4 w( h) = 5 h 9 We find that w( h) = 4 ( 5 h) 9 According to ( ) , the total force on the side is then
4 h 624 ( 5 h) dh 577778 lbs 9
EXAMPLE
An aquarium tank is filled with a mixture of water and algicide to keep the liquid clear for viewing The liquid has a density of 50 pounds per cubic foot For viewing purposes, a window is located in the side of the tank, with center 20 feet below the surface The window is in the shape of a square of side 4 2 feet with vertical and horizontal diagonals (see Figure 841) What is the total force on this window SOLUTION
As usual, we measure depth downward from the surface with independent variable h Then the range of integration will be h = 20 4 = 16 to h = 20 + 4 = 24 Refer to Figure 842 For h between 16 and 20, we notice that the right triangle in Figure 842 is isosceles and hence has base of length 8 A P P L I C A T I O N S O F T H E I N T E G R A L
FIGURE 841
h 16 Therefore w( h) = 2( h 16) = 2h 32 According to our analysis, the total force on the upper half of the window is thus 20 16 h 50 ( 2h 32) dh =
44880 lbs 3
For the lower half of the window, we examine the isosceles right triangle in Figure 843 It has base 24 h Therefore, for h ranging from 20 to 24, we have w( h) = 2( 24 h) = 48 2h h = 16 4 2 h _ 16 h = 24 FIGURE 842
CALCULUS DeMYSTiFieD
h = 16 4 2 24 _ h
h = 24 FIGURE 843
According to our analysis, the total force on the lower half of the window is
24 20 h 50 ( 48 2h) dh =
51200 lbs 3
The total force on the entire window is thus 44880 51200 96080 + = lbs 3 3 3
YOU TRY IT A tank of water has flat sides On one side, with center 4 feet below
the surface of the water, is a circular window of radius 1 foot What is the total force on the window 87 Numerical Methods of Integration
While there are many integrals that we can calculate explicitly, there are many others that we cannot For example, it is impossible to evaluate e x dx 8 A P P L I C A T I O N S O F T H E I N T E G R A L
That is to say, it can be proved mathematically that no closedform antideriva2 tive can be written down for the function e x Nevertheless, ( ) is one of the most important integrals in all of mathematics, for it is the Gaussian probability distribution integral that plays such an important role in statistics and probability Thus we need other methods for getting our hands on the value of an integral One method would be to return to the original definition, that is to the Riemann sums If we need to know the value of

