Geomath Geology 351 - Final Review - Part 1 tom.h.wilson tom.wilson@mail.wvu.edu Dept. Geology and Geography West Virginia University

Evaluating those definite integrals 9.9 & 9.10 Evaluating those definite integrals

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

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

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

The “definite” solution

Strain Li Lf If we consider two deformations L1 and L2, the final strain is different from the sum of the strains

The sum is expressed as a definite integral The true strain or total natural strain, can be totaled. It is the sum of an infinite number of infinitely small extensions and is also known as the logarithmic strain The sum is expressed as a definite integral Where s is known as the stretch ratio or relative elongation and is also denoted as 

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

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

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

The heat generated will be This is a differential quantity so there is no need to integrate iii & iv. Calculate total 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.

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

h=85feet IN-CLASS PROBLEMS 1. At the base of a cliff you are standing on top of geological Unit A. The cliff face is formed along a normal fault (nearly vertical). The top of Unit A is also exposed at the top of the cliff face. You walk a distance x = 200 feet away from the fault scarp. Looking back toward the cliff, you use your Brunton and measure and note that the top of the cliff is 23o above the horizon. What is the offset along this fault? h x  Cliff Face Top of Unit A h=85feet

3. In the example illustrated below, a stream erodes less resistant fault gauge leaving an exposed fault scarp on the distant bank. You are unable to traverse the stream or make your way to the top of the exposure. Using your Brunton compass, you stand on the left edge of the stream and measure the angle (a) formed by the top of the cliff and the horizontal. You walk to the left 175 feet and measure angle (b). Angle a measure 31o and angle b, 19o. How can you determine the cliff height? What is the width of the stream? h d a b Stream Fault Scarp h~230 feet

4. The three point problem uses elevations measured at three points on a stratigraphic surface to determine the strike and dip of that surface. The elevations and locations of these points can be measured at the surface or, more likely, in the borehole. In the following problem, you have data from three boreholes (located in the map below) indicating subsea depths to the top of the Oriskany Sandstone as shown. Determine the strike and dip. -4300’ -4700’ -5000’ N Scale: 1:10000 ~N50W ~24SW

General overview of the 3 point problem

 = N56W

Now, how can you determine the dip?  2080 feet 300 feet Measure length of line 3. Given this length and the drop in elevation you can figure the dip  directly -

Remember vector addition and subtraction begin by breaking the vectors down into their x & y (and sometimes z) components. You can do this graphically or trigonometrically.

Trig breakdown

Now, how would you calculate the dip? 3500 3000 2000 4000 =N69W ~625’ 3500 3000 2500 2500 100 feet

Questions of this type will not be covered on the exam. The strike on the sandstone boundary of 72o implies an  of 32o (angle cliff face makes with actual dip direction). The apparent dip of the beds along the cliff face is The true dip is

Continue your review Look over example test questions in addition to the material we covered today and bring your questions to class on Thursday