Dec 30, 2015
3 2
03 2 1
xax bx cx
2
0
xax bx c dx
For the 5th order polynomial you derive you’ll have 6 terms including the constant
Some useful calculus linkshttp://
archives.math.utk.edu/visual.calculus/http://archives.math.utk.edu/utk.calculus/142toc.html
http://archives.math.utk.edu/visual.calculus/4/substitutions.3/index.html
Integration by substitution- a quick
illustration2 343 8 3t t dt
The problem we illustrated in class the other day is a little complicated
A lot of the problems on the in-class worksheet are similar to this. To simplify these kinds of problems, you look for something that when differentiated supplies another term in the integrand.
In this problem, we can see that 3
28 3
9d t
tdt
which within a factor of 1/3rd equals the leading term in the integrand – the 3t2. So the idea is that we define a new variable
3
3 3
8 3 , where
9 and 9
u t
dut du t dt
dt
2 343 8 3t t dt
We also see that 33
3
dut dt
So we substitute our new terms 33
3
dut dt38 3u t
into
to get 1 14 4
1
3 3
duu u du
14
1
3u duNow you have the much simpler
integral to evaluate.You just need to use the
power rule on this.
What would you get? 5
41 4
3 5u
Substitute the expression defined as
u(t) back in
What is the volume of Mt. Fuji?
1
N
ii
V V
2
1
N
ii
r z
max
min
2Z
iZV r dz
The volume of each of the little disks (see below) represents a
iV
2i i iV r z
max
min
2Z
iZV r dz
2ir dz
ridz
is the volume of a disk having radius r and thickness dz.
=total volumeThe sum of all disks with
thickness dz
Area
riRadius
2 2400 800400
3 3
z zr km
3 2
0
400 800400
3 3
z zV km
3 3 3
0 0 0
400 800400
3 3
z zV dz dz dz
Waltham notes that for Mt. Fuji, r2 can be approximated by the following polynomial
To find the volume we evaluate the definite integral
Li
Lfi
Ls
L
f
i
L
L
dL
L ln f
i
L
LL
ln( ) ln( )
ln
f i
f
i
L L
L
L
ln( )S
The total natural strain, , is the sum of an infinite number of infinitely small extensions
In our example, this gives us the
definite integral
11,000 kg/m3
We can simplify the problem and still obtain a useful result. Approximate the average
densities
4,500 kg/m3
2
1
4N
i ii
M r r
2
04
RM r dr
11,000 kg/m3
We can simplify the problem and still obtain a useful result. Approximate the average
densities
4,500 kg/m3
3480 63712 2
0 34804 .11000 4 .4500r dr r dr
2
04
RM r dr
34
3r C
3480 63713 3
1 20 3480
4 4
3 3r r
3480 63713 3
20 3480
44000 18000
3 3r r
The result – 6.02 x 1024kg is close to the generally accepted
value of 5.97 x 1024kg.
320
y kWQ
km
3
00
20
kW kW
km km 3 3
100.5
20
kW kW
km km
Where y is the distance in km from the base of the Earth’s crust
i. Determine the heat generation rate at 0, 10, 20, and 30 km from the base of the crust
V A z 3V zkm
ii. What is the heat generated in a in a small box-shaped volume z thick and 1km x 1km surface? Since
1
n
ii
q Q z
0
0
201
20
z
z
zq dz
zdz
The heat generated will be 3Q V Q zkm
iii & iv. Heat generated in the vertical column
This is a differential quantity so there is no need to integrate
In this case the sum extends over a large range of z, so However, integration is the way to go.
0
zq Qdz
302
0
22.540
yq kW
v. Determining the flow rate at the surface would require evaluation of the definite integral
vi. To generate 100MW of power
2
2
100,0004444
22.5
kWkm
kWkm