Download Applications of Integration

Applications of
Integration
Section 7.5b
Fluid Force and Fluid Pressure
We build dams wider at the bottom than at the top.
Why?
What does the greater pressure at the bottom of the
dam depend upon?
 Only on the type of liquid and how far below the
surface the point lies (not on how much liquid the
dam is holding back)
Read about “The Great Molasses Flood of 1919” at
the top of p.404…
Fluid Force and Fluid Pressure
(a) Given that the tank was full of molasses weighing
100 lb ft 3, what was the total force exerted by the molasses
on the bottom of the tank at the time it ruptured?
In any liquid, the fluid pressure p (force per unit area)
at depth h is
p  wh
where w is the weight-density (weight per unit volume of
the given liquid)
Fluid Force and Fluid Pressure
(a) Given that the tank was full of molasses weighing
100 lb ft 3, what was the total force exerted by the molasses
on the bottom of the tank at the time it ruptured?
At the bottom of this tank, the pressure from the molasses:
lb
lb 

p  wh  100 3   90ft   9000 2
ft
ft 

2
The area of the base of the tank:   45   2025
Total force on the base:
lb 

2
 9000 2   2025 ft   57, 255,526 lb
ft 

Fluid Force and Fluid Pressure
(b) What was the total force against the bottom foot-wide
band of the tank wall?
Partition the band into narrower bands of width y, depth
Pressure at this depth:
p  wh  100 yk
Force against each narrow band = (pressure)(area):
100 yk   2  45 y   9000 yk y
lb
yk
Fluid Force and Fluid Pressure
(b) What was the total force against the bottom foot-wide
band of the tank wall?
Force against each narrow band = (pressure)(area):
100 yk   2  45 y   9000 yk y
lb
Summing these forces for the entire partition:
90
y 
F   9000 ydy  9000  
89
 2  89
7921 

 9000  4050 
  805,500
2   2,530,553 lb

90
2
Fluid Force and Fluid Pressure
A rectangular milk carton measures 3.75in by 3.75in at the
base and is 7.75in tall. Find the force of the milk (weighing
64.5lb ft 3 ) on one side when the carton is full.
3.75in  5 16 ft
7.75in  31 48ft
Force equals pressure applied over an area:
p  64.5 y
F 
31 48
0
A   5 16  dy
31 48
 5  645  y 
 64.5 y   dy  


 16  32  2  0
2
 4.204 lb
Normal Probabilities
In mathematics, probabilities are represented as areas, which
is where integrals come into play…
Definition: Probability Density Function (pdf)
A probability density function is a function f  x  with
domain all reals such that
f  x   0 for all x
and



f  x  dx  1
Then the probability associated with an interval  a, b  is
 f  x  dx
b
a
Normal Probabilities
By far the most useful pdf is the normal kind…
Definition: Normal Probability Density Function (pdf)
A normal probability density function (Gaussian curve)
for a population with mean  and standard deviation  is
 x     2 2 
1
f  x 
e
 2
2
Normal Probabilities
And an extremely valuable rule…
The 68-95-99.7 Rule for Normal Distributions
Given a normal curve,
 of the mean 
95% of the area will lie within 2 of the mean 
99.7% of the area will lie within 3 of the mean 
• 68% of the area will lie within
•
•
Normal Probabilities
Suppose that frozen spinach boxes marked as “10 ounces” of
spinach have a mean weight of 10.3 ounces and a standard
deviation of 0.2 ounce.
The pdf:
f  x 
1
0.2 2
e
 x 10.3
2
Graph this function in [9, 11.5] by [-1, 2.5]
Does the function have the expected shape?
 0.08
Normal Probabilities
Suppose that frozen spinach boxes marked as “10 ounces” of
spinach have a mean weight of 10.3 ounces and a standard
deviation of 0.2 ounce.
(a) What percentage of all such spinach boxes can be
expected to weigh between 10 and 11 ounces?
The probability that a randomly chosen spinach box weighs
between 10 and 11 ounces is the area under the curve from
10 to 11. Find numerically:
NINT  f  x  , x,10,11  0.93296
There is approximately a 93.296% chance that a box
will weigh between 10 and 11 ounces.
Normal Probabilities
Suppose that frozen spinach boxes marked as “10 ounces” of
spinach have a mean weight of 10.3 ounces and a standard
deviation of 0.2 ounce.
(b) What percentage would we expect to weigh less than
10 ounces?
We need the area under the curve to the left of x = 10…
Because of the asymptote, we use x = 9 as an approximation
for the lower bound:
NINT  f  x  , x,9,10   0.06681
There is approximately a 6.681% chance that
a box will weigh less than 10 ounces.
Normal Probabilities
Suppose that frozen spinach boxes marked as “10 ounces” of
spinach have a mean weight of 10.3 ounces and a standard
deviation of 0.2 ounce.
(c) What is the probability that a box weight exactly 10
ounces?
This would be the integral from 10 to 10, which is zero.
Huh???
Remember that we are assuming a continuous, unbroken
interval of possible weights, and 10 is but one of an infinite
number of them…