Applications of IntegrationSection 7.5b

Fluid Force and Fluid Pressure

Read about “The Great Molasses Flood of 1919” atthe top of p.404…

We build dams wider at the bottom than at the top.Why?

What does the greater pressure at the bottom of thedam depend upon?

Only on the type of liquid and how far below thesurface the point lies (not on how much liquid thedam is holding back)

Fluid Force and Fluid Pressure(a) Given that the tank was full of molasses weighing , what was the total force exerted by the molasseson the bottom of the tank at the time it ruptured?

3100lb ft

In any liquid, the fluid pressure p (force per unit area)at depth h is p whwhere w is the weight-density (weight per unit volume ofthe given liquid)

Fluid Force and Fluid Pressure(a) Given that the tank was full of molasses weighing , what was the total force exerted by the molasseson the bottom of the tank at the time it ruptured?

3100lb ft

At the bottom of this tank, the pressure from the molasses:

3

lb100 90ft

ftp wh

2

lb9000

ft

The area of the base of the tank: 245 2025

Total force on the base:

22

lb9000 2025 ft

ft

57,255,526 lb

Fluid Force and Fluid Pressure(b) What was the total force against the bottom foot-wideband of the tank wall?

Partition the band into narrower bands of width , depthy ky

Pressure at this depth: 100 kp wh y

Force against each narrow band = (pressure)(area):

100 2 45ky y 9000 ky y lb

Fluid Force and Fluid Pressure(b) What was the total force against the bottom foot-wideband of the tank wall?

Force against each narrow band = (pressure)(area):

100 2 45ky y 9000 ky y lb

Summing these forces for the entire partition:

90

899000F ydy

902

89

90002

y

79219000 4050

2

805,500

2,530,553 lb

Fluid Force and Fluid PressureA rectangular milk carton measures 3.75in by 3.75in at thebase and is 7.75in tall. Find the force of the milk (weighing ) on one side when the carton is full.364.5lb ft

3.75in 5 16ft 7.75in 31 48ftForce equals pressure applied over an area:

64.5p y 5 16A dy31 482

0

645

32 2

y

31 48

0

564.5

16F y dy

4.204 lb

Normal ProbabilitiesIn mathematics, probabilities are represented as areas, whichis where integrals come into play…

Definition: Probability Density Function (pdf)

A probability density function is a function withdomain all reals such that

f x

0f x for all x and 1f x dx

Then the probability associated with an interval is ,a b

b

af x dx

Normal ProbabilitiesBy 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

2 221

2

xf x e

Normal ProbabilitiesAnd an extremely valuable rule…

The 68-95-99.7 Rule for Normal Distributions

Given a normal curve,

• 68% of the area will lie within of the mean • 95% of the area will lie within of the mean2 • 99.7% of the area will lie within of the mean3

Normal ProbabilitiesSuppose that frozen spinach boxes marked as “10 ounces” ofspinach have a mean weight of 10.3 ounces and a standarddeviation of 0.2 ounce.

Does the function have the expected shape?

The pdf: 210.3 0.081

0.2 2xf x e

Graph this function in [9, 11.5] by [-1, 2.5]

Normal ProbabilitiesSuppose that frozen spinach boxes marked as “10 ounces” ofspinach have a mean weight of 10.3 ounces and a standarddeviation of 0.2 ounce.

(a) What percentage of all such spinach boxes can beexpected to weigh between 10 and 11 ounces?

The probability that a randomly chosen spinach box weighsbetween 10 and 11 ounces is the area under the curve from10 to 11. Find numerically:

NINT , ,10,11f x x 0.93296There is approximately a 93.296% chance that a box

will weigh between 10 and 11 ounces.

Normal ProbabilitiesSuppose that frozen spinach boxes marked as “10 ounces” ofspinach have a mean weight of 10.3 ounces and a standarddeviation of 0.2 ounce.

(b) What percentage would we expect to weigh less than10 ounces?

We need the area under the curve to the left of x = 10…

NINT , ,9,10f x x 0.06681There is approximately a 6.681% chance that

a box will weigh less than 10 ounces.

Because of the asymptote, we use x = 9 as an approximationfor the lower bound:

Normal ProbabilitiesSuppose that frozen spinach boxes marked as “10 ounces” ofspinach have a mean weight of 10.3 ounces and a standarddeviation of 0.2 ounce.

(c) What is the probability that a box weight exactly 10ounces?

This would be the integral from 10 to 10, which is zero.

Huh???Remember that we are assuming a continuous, unbrokeninterval of possible weights, and 10 is but one of an infinitenumber of them…