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Geomath

Geomath. Geology 351 -. Calculus VI. tom.h.wilson tom.wilson@mail.wvu.edu. Dept. Geology and Geography West Virginia University. Example 9.7 - find the cross sectional area of a sedimentary deposit (see handout). Set-up.

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Geomath

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  1. Geomath Geology 351 - Calculus VI tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

  2. Example 9.7 - find the cross sectional area of a sedimentary deposit (see handout).

  3. Set-up For the 5th order polynomial you derive you’ll have 6 terms including the constant

  4. Example Set-up: 4th order 4th order example from the text t = -2.857E-12x4 + 1.303E-08x3 - 2.173E-05x2 + 1.423E-02x - 7.784E-02

  5. Some useful calculus links http://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

  6. Integration by substitution- a quick illustration 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.

  7. In this problem, we can see that 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

  8. We also see that So we substitute our new terms & into to get

  9. Now you have the much simpler integral to evaluate. You just need to use the power rule on this. What would you get? Substitute the expression defined as u(t) back in

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

  11. in-class integration practice How did you do?

  12. Volumes too ... What is the volume of Mt. Fuji? The volume of each of the little disks (see below) represents a

  13. ri dz Radius ri is the volume of a disk having radius r and thickness dz. Area =total volume The sum of all disks with thickness dz

  14. Waltham notes that for Mt. Fuji, r2 can be approximated by the following polynomial To find the volume we evaluate the definite integral

  15. The “definite” solution

  16. Li Strain Lf The total natural strain, , is the sum of an infinite number of infinitely small extensions In our example, this gives us the definite integral

  17. We can simplify the problem and still obtain a useful result. Approximate the average densities 11,000 kg/m3 4,500 kg/m3 discontinuous Integrating Functions

  18. We can simplify the problem and still obtain a useful result. Approximate the average densities 11,000 kg/m3 4,500 kg/m3 The result – 6.02 x 1024kg is close to the generally accepted value of 5.97 x 1024kg.

  19. 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 ii. What is the heat generated in a in a small box-shaped volume z thick and 1km x 1km surface? Since

  20. The heat generated will be This is a differential quantity so there is no need to integrate iii & iv. Heat generated in the vertical column In this case the sum extends over a large range of z, so However, integration is the way to go.

  21. v. Determining the flow rate at the surface would require evaluation of the definite integral vi. To generate 100MW of power

  22. Questions? Problems 9.9 and 9.10?

  23. Hand integral worksheets in before leaving • Finish up problem 9.7 for Tuesday • 3. We’ll start our review of trigonometry on Tuesday, so look over Chapter 5.

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