Integration - mathdoctor1999.com

advertisement

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 Integration

Who 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…

Work

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

Think: What do I

It will be slightly modified. remember about work?

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].

Question

How 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:  x

 b

 n a

. Thus we divide the interval [a, b] up into n subintervals each of length

 x

 x i

.

Connect

x i i x i

 x i i n 

1

F

  i

 x i

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

x i

Conclude

Thus lim n

  i

N 

1

F

  i

 x i is the definition of work .

Since F is continuous, the limit is the integral x x

 b a

F

  dx which is the definition of work.

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.

Strategy

Visualize: the problem.

Who wants to volunteer to draw a picture?

Diameter of tank is 6 feet

The Picture

Lowest point of tank is 12 feet from ground.

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

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?

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.

Moving layer of water to a height of 15 - y

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.

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

layer of water?

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

F

 weight



62 lbs ft

3



 volume

62

 x

2  y

 pounds

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 x

2 y

2 can use the formula for a circle to find in terms of .

x

2 

 y

15

2 

3

2 

9 x

2   y

2 

30 y

216

F

62

Thus the increment of force can be written as:

 x

2  y

 pounds

62

 y

2 

30 y

216

 y

Thus the increment of work can be written now as:

W

62

F y

3 

15

 y

45 y

2 

62

666 y

 y

2 

3240

30

 y y

216

 y

15

 y

Integrate

Compute: Take all the information and formulate the integral then integrate.

W

15

12

62

 y

3 

45 y

2 

666 y

3240

 dy

62

 y

4

4

15 y

3 

333 y

2 

3240 y

15

12

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

Center (0,3) and radius of 3

Sitting on the x-axis

Download