Geomath

tom.h.wilsontom.wilson@mail.wvu.

eduDept. Geology and

GeographyWest Virginia University

Example 9.7 - find the cross sectional area of a sedimentary deposit (see

handout).

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

t = -2.857E-12x4 + 1.303E-08x3 - 2.173E-05x2 + 1.423E-02x - 7.784E-

02

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

5

3 448 3

15tTo get

Integration by substitution helps structure the process of finding a solution.

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

32 1.5

0

400 800400

6 1.5 3

z zz

600 1600 1200

3200 628km

3 2

0 iV r dz

The “definite” solution

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

1. Hand integral worksheets in before leaving

2. Finish up problem 9.7 for Tuesday

3. We’ll start our review of trigonometry on Tuesday, so look over Chapter 5.