Transcript Integration - mathdoctor1999.com

Integration
Work as an Application
The BIG Question
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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?
It will be slightly
Think: What do I
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a
.
n
Thus we divide
the interval [a, b] up into n subintervals each of length x
 xi
Connect
Connect: So, let x i be some point in the subinterval so our constant for F is F(x i ).
Conclude: The work on the i th subinterval is F(x i ) xi
.
Connect: Totaling the work on each subinterval we have
n
 F x. x
i 1
i
Think: If I let N get larger, the partition gets finer, and I get a
more accurate estimate of work.
i
Conclude
N
Thus
lim  F x x
n 
i 1
i
i
is the definition of work.
Since F is continuous, the limit is the integral
x b
 F x dx
x a
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?
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?
Layer of water
of thickness
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.
y
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.
Question: What is the increment of force or  F for this
layer of water?
Connect: The increment of force is determined by the weight
of the layer which we have as:
 62lbs 
F  weight   3 volume  62 x 2 y pounds
 ft 
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
2
2
can use the formula for a circle to find x in terms of y .
x 2   y  15   32  9
2
x 2   y 2  30 y  216
Thus the increment of force can be written as:




F  62 x 2 y pounds 62  y 2  30y  216 y
Thus the increment of work can be written now as:


W  F 15  y   62  y 2  30 y  216 y15  y 


 62 y 3  45y 2  666y  3240 y
Integrate
Compute: Take all the information and formulate
the integral then integrate.
15
 y4

3
2
W   62 y  45y  666y  3240 dy  62   15y 3  333y 2  3240y 
4
12
12
15


= 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
Center (0,3) and
radius of 3
Sitting on the
x-axis