Integration

Work as an Application

The BIG Question

Did you prepare for today?

If so, estimate the time you spent preparing and write it down on your frequency log for today.

Definition of IntegrationWho recalls what integration is?

The definite integral is informally defined to be the area of the region in the xy-plane bounded by the graph of ƒ(x), the x-axis, and the vertical lines x = a and x = b. We use partitioning to write the area as a sum of the areas of rectangles. As the partition becomes more fine, the…

WorkDo you think the definition of integral will change if we use a definite integral to calculate work?

It will be slightly modified.

How? What is the definition of work using integrals?

If we have a constant force, F, moving an object d distance along a straight line, recall W = F•d.

Connect: Suppose we have that F is a continuous function on a closed interval [a, b].

Think: What do I remember about work?

QuestionHow do we define work using integrals moving an object from x = a to x = b?

Think: We need to find a partition to define the work done along the interval [a, b].

Connect: Let n be a positive number so that: . Thus we divide

the interval [a, b] up into n subintervals each of length x

.

n

abx

ix

ConnectConnect: So, let be some point in the subinterval so our constant for F is F( ).

Conclude: The work on the subinterval is F( ) .

Connect: Totaling the work on each subinterval we have . i

n

ii xxF

1

thi

Think: If I let N get larger, the partition gets finer, and I get a more accurate estimate of work.

ix

ix ix

ix

Conclude

Thus is the definition of work.

bx

ax

dxxF

Since F is continuous, the limit is the integral

which is the definition of work.

i

N

ii

n

xxF 1

lim

Example 1: On the densely populated island of Okinawa, water shortages are common, and homes are typically equipped with a rooftop water tank in the shape of a cylinder or sphere.

Problem:One spherical tank with radius 3 feet is mounted so its lowest point is 12 feet above ground. How much work is done in pumping water from ground level to fill the tank half way? Water weighs about 62 lbs per cubic foot.

StrategyVisualize: the problem.

Who wants to volunteer to draw a picture?

The Picture

Diameter of tank is 6 feet

But only fill half way with water up to 3 feet.

Lowest point of tank is 12 feet from ground.

Groundwork

Question: How can I use integration to define the amount of work needed to pump the water from ground level to a level of 3 feet in the tank?  

Connect: I need to know how W = F•d connects with this problem. I also need to know my interval [a, b]. However, I am moving it from 12ft. to half way up or to 15ft. So the interval is [12, 15]. Here we have a circle with center of (0,15) with a radius of 3.

Connect: The information given to the definition of the things we need to define for the integral. I define a layer of water with thickness of a distance (15 - y) feet from the bottom of the tank.

Picture It

Any volunteers again?

Moving layer of water to a

height of 15 - y

Layer of water of thickness

y

Here we have a center of (0,15) with a radius of 3

This layer does not have radius of 3, but has radius of x.

Connect Using this information, I calculate the weight of this one layer of

water, then sum up the weights of all layers using an integral.

Question: What is the increment of force or for thislayer of water?

F

Connect: The increment of force is determined by the weight of the layer which we have as:

This layer does not have radius of 3, but has radius of x.

poundsyxvolumeft

lbsweightF

2

362

62 Volume is surface area multiplied by thickness of the layer of water.

Make Equation

NOW, since the sphere has radius of 3 ft and for a circle with center at (0, 15), we can use the formula for a circle to find in terms of .

2x 2y

21630

931522

222

yyx

yx

Thus the increment of force can be written as:

yyypoundsyxF 216306262 22

Thus the increment of work can be written now as:

yyyy

yyyyyFW

32406664562

1521630621523

2

IntegrateCompute: Take all the information and formulate

the integral then integrate.

15

12

15

12

234

23 3240333154

6232406664562 yyyy

dyyyyW

= 62 [20.25] = 1255.5

Summarize: The work needed to fill a spherical tank of radius 3ft. half way with the lowest point 12ft. from the ground to a level of 15ft. from the ground is W = 1255.5 foot-pounds

You Try:

Is there an alternate way to set up and solve this equation?

YES!!!

Hint: Set the bottom of the spherical tank on the x-axis with center at the point (0,3).

• How does this simplify the problem? • What would the center and radius be if we did the problem this

way?

New Picture

Sitting on the x-axis

Center (0,3) and radius of 3