1. Tom Wilson, Department of Geology and Geography tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

2. Any questions about integration problems 1-4? Tom Wilson, Department of Geology and Geography

3. Let’s continue on with these (the last set of general integrals for you to solve) Tom Wilson, Department of Geology and Geography

4. Another set Tom Wilson, Department of Geology and Geography

5. Last set Tom Wilson, Department of Geology and Geography

6. A brief introduction to integration by substitution Tom Wilson, Department of Geology and Geography Take a look at this one for a minute. 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. Tom Wilson, Department of Geology and Geography In this problem, we can see that which within a factor of 1/3 rd equals the leading term in the integrand – the 3t 2. So the idea is that we define a new variable

8. Tom Wilson, Department of Geology and Geography We also see that So we substitute and redefine the integral into to get

9. Tom Wilson, Department of Geology and Geography 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. Tom Wilson, Department of Geology and Geography Our result … Integration by substitution helps structure the process of finding a solution. Differentiate to verify.

11. Try another one Tom Wilson, Department of Geology and Geography Let u =

12. Let’s consider some heat flow problems as a companion discussion to the example in Chapter 9 Tom Wilson, Department of Geology and Geography Consider heat conduction through a thick glass window given two possible inside temperatures 65 o F and 72.2 o F and an outside temperature of 32 o F. In terms of degrees C this corresponds to temperatures of 18.33 O C, 22.33 o C and 0 o C. How much energy do you save? See http://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html for additional discussionhttp://serc.carleton.edu/NAGTWorkshops/geophysics/activities/18913.html This problem is solved using a simple equation referred to as the heat conduction equation.

13. Tom Wilson, Department of Geology and Geography We consider this problem in terms of the heat flow over the course of the day, where heat flow (q x ) is expressed in various units representing heat per unit area per time: for example, calories/(m 2 -s). A q x of 1 cal/(cm 2 -s)=41.67 kW/m 2 kW-s=737.5622 ft-lbs 1 calorie/sec = 0.004284 kW =4.1868 Watts

14. Relating to the units Tom Wilson, Department of Geology and Geography If you lift about 3 pounds one foot in one second, then you’ve expended 1 calorie (thermomechanical) of energy. Nutritional calories are about 1000 thermomechanical calories. So you would have burnt only 1/1000th a nutritional calorie! You can expend 1 nutritional calorie by carrying 100 lbs up 30 feet. or =3.086 ft-lbs/second

15. To solve this problem, we use the heat conduction equation: q x =-K  T/  x Tom Wilson, Department of Geology and Geography K (thermal conductivity)  2x10 -3 cal/(cm-sec- o C) Assume  x=0.5cm Then q x =0.07 cal/(cm 2 -sec) or 0.089 cal/(cm 2 -sec) If the window has an area of 2m 2 Then the net heat flowing across the window is 733 or 896 cal/sec The lower temperature saves you 163 cal/sec

16. 80,000 cal/sec Tom Wilson, Department of Geology and Geography 163 cal/sec corresponds to 1.41x10 7 cal/day There are 860420.650 cal per kWh so that this corresponds to about 16.3 kWh/day. A kWh goes for about 6.64 cents so $1.08/day. note this estimate depends on an accurate estimate of K (thermal conductiviry). Other values are possible, but, as you can see, it can add up!

17. The second question concerns a hot sill Tom Wilson, Department of Geology and Geography A hot sill intruded during Mesozoic time is now characterized by temperature from east-to-west that varies as X=0 km X=40 km

18. What is the derivative Tom Wilson, Department of Geology and Geography You see you have to take a derivative to determine heat flow. Using the conduction equation in differential form

19. Calculate the temperature gradient Tom Wilson, Department of Geology and Geography Given K and that 1 heat flow unit = Calculate q x at x=0 and 40km.

20. and substitute for x to get q Tom Wilson, Department of Geology and Geography You will get q x =-1.8 hfu at x=0 and q x =0.6 hfu at x=40 X=0 km X=40 km Heat flows out both ends of the sill.

21. Another simple example : assume the mantle and core have the same heat production rate as the crust Tom Wilson, Department of Geology and Geography What is the heat flow produced by the Earth in this case? Is it a good assumption? Typical radiogenic heat production  for granite is ~2x10 -13 cal/(gm-sec) and that for basalt – about 2x10 - 14. We use an average of about 1x10 -13 for this problem. Given that the mass of the Earth is about 6x10 27 grams we get a heat generation rate of about 6.6x10 14 cal/sec. What is the heat flow per cm 2 ?

22. Heat flow per unit area … Tom Wilson, Department of Geology and Geography To answer that, we need the total area of the Earth’s surface in cm 2. The surface area of the Earth is about 5.1 x 10 18 cm 2. which gives us a heat generation rate /cm 2 of about 12.9 x10 -5 cal/(cm 2 -sec) or 129 hfu. The global average heat flow is about 1.5 hfu. We would have to conclude that the earth does not get much radiogenic heating from the mantle and core.

23. For Thursday Tom Wilson, Department of Geology and Geography Review problem 9.8. In this problem, we assume that heat is generated in the crust at the rate of 1kW/km 3 and that heat generation is confined to the crust (we’ve confirmed this is a pretty good assumption). Radiogenic heating decreases with depth until, as we suspected, below the crust there is very little heat generated through radioactive decay (the thermal gradient or direction of heat flow is from hot to cold or vertically upward).

24. Problem 9.8 Tom Wilson, Department of Geology and Geography Heat generation rate in this problem is defined as a function distance from the base of the crust. Waltham uses y for this variable and expresses heat generation rate (Q) as Also review total natural strain discussion and the integration of discontinuous functions.

25. Return to example 9.7 where your task is to find the cross sectional area of the sand body Tom Wilson, Department of Geology and Geography

26. Waltham presents the results from a 4 th order polynomial approximation of the sand bar thickness Tom Wilson, Department of Geology and Geography t = -2.857E-12x 4 + 1.303E-08x 3 - 2.173E-05x 2 + 1.423E-02x - 7.784E-02 Recall how to compute the cross sectional area?

27. In this exercise we repeat Wlatham’s analysis using a 5 th order polynomial Tom Wilson, Department of Geology and Geography For the 5 th order polynomial you derive you’ll have 6 terms including the constant What is this integral?

28. Recall computation set up using the 2 nd order polynomial as an example Tom Wilson, Department of Geology and Geography The integral has factors a/3, b/2 and c. The values a, b, and c come from the best-fit process. You already know what those numbers are from the trendline. You also know the limits of integration: 0 to 2000meters

29. Tom Wilson, Department of Geology and Geography For the 5 th order polynomial you derive you’ll have 6 terms. Enter values here (Upper limit of x) 6 (Upper limit of x) 5 Etc.

30. Tom Wilson, Department of Geology and Geography For the 5 th order polynomial you derive you’ll have 6 terms.

31. Problem will be due next Tuesday Tom Wilson, Department of Geology and Geography We’ll give you an extra day on this, so bring questions to class on Thursday.

32. Tom Wilson, Department of Geology and Geography 1.Hand the integral worksheets in before leaving or put in my mailbox sometime today. 2.Finish up problem 9.7 for next Tuesday 3.We’ll review other problems in the text this Thursday (e.g. 9.8 & discussions of topics illustrated in figures 9.4, 9.5 & 9.6). 4. Also look over problems 9.9 and 9.10 and bring questions to class this Thursday.