USING THE DEFINITE INTEGRAL

Print this page Chapter Eight USING THE DEFINITE INTEGRAL 8.1 AREAS AND VOLUMES In Chapter 5, we calculated areas under graphs using definite integrals. We obtained the integral by slicing up the region, constructing a Riemann sum, and then taking a limit. In this section, we calculate areas of other regions, as well as volumes, using definite integrals. To obtain the integral, we again slice up the region and construct a Riemann sum. Finding Areas by Slicing Example 1 Use horizontal slices to set up a definite integral to calculate the area of the isosceles triangle in Figure 8.1. Figure 8.1 Isosceles triangle Solution Notice that we can find the area of a triangle without using an integral; we will use this to check the result from integration: To calculate the area using horizontal slices we divide the region into strips; see Figure 8.2. A typical strip is approximately a rectangle of length w i and width Δ h, so Area of strip w i Δ h cm 2. Figure 8.2 Horizontal slicing of isosceles triangle Page 1 of 17

To get w i in terms of h i, the height above the base, use the similar triangles in Figure 8.2: Summing the areas of the strips gives the Riemann sum approximation: Taking the limit as n, the change in h shrinks and we get the integral: Evaluating the integral gives Notice that the limits in the definite integral are the limits for the variable h. Once we decide to slice the triangle horizontally, we know that a typical slice has thickness Δ h, so h is the variable in our definite integral, and the limits must be values of h. Example 2 Use horizontal slices to set up a definite integral representing the area of the semicircle of radius 7 cm in Figure 8.3. Figure 8.3 Semicircle Solution As in Example 1, to calculate the area using horizontal slices, we divide the region into strips; see Figure 8.4. A typical strip at height h i above the base has width w i and thickness Δ h, so Area of strip w i Δ h cm 2. Figure 8.4 Horizontal slices of semicircle Page 2 of 17

To get w i in terms of h i, we use the Pythagorean Theorem in Figure 8.4: so Summing the areas of the strips gives the Riemann sum approximation Taking the limit as n, the change in h shrinks and we get the integral: Using the table of integrals VI-30 and VI-28, or a calculator or computer, gives As a check, notice that the area of the whole circle of radius 7 is π 7 2 = 49π cm 2. Finding Volumes of Slicing To calculate the volume of a solid using Riemann sums, we chop the solid into slices whose volumes we can estimate. Let's see how we might slice a cone standing with the vertex uppermost. We could divide the cone vertically into arch-shaped slices; see Figure 8.5. We could also divide the cone horizontally, giving coin-shaped slices; see Figure 8.6. Figure 8.5 Cone cut into vertical slices Page 3 of 17

Figure 8.6 Cone cut into horizontal slices To calculate the volume of the cone, we choose the circular slices because it is easier to estimate the volumes of the coin-shaped slices. Example 3 Use horizontal slicing to find the volume of the cone in Figure 8.7. Figure 8.7 Cone Solution Each slice is a circular disk of thickness Δ h. See Figure 8.7. The disk at height h i above the base has radius. From Figure 8.8 and the previous example, we have w i = 10 2h i so r i = 5 h i. Figure 8.8 Vertical cross-section of cone in Figure 8.7 Each slice is approximately a cylinder of radius r i and thickness Δ h, so Summing over all slices, we have Page 4 of 17

Taking the limit as n, so Δ h 0, gives The integral can be evaluated using the substitution u = 5 h or by multiplying out (5 h ) 2. Using the substitution, we have Note that the sum represented by the sign is over all the strips. To simplify the notation, in the future, we will not write limits for or subscripts on the variable, since all we want is the final expression for the definite integral. We now calculate the volume of a hemisphere by slicing. Example 4 Set up and evaluate an integral giving the volume of the hemisphere of radius 7 cm in Figure 8.9. Figure 8.9 Slicing to find the volume of a hemisphere Solution We will not use the formula for the volume of a sphere. However, our approach can be used to derive that formula. Divide the hemisphere into horizontal slices of thickness Δ h cm. (See Figure 8.9.) Each slice is circular. Let r be the radius of the slice at height h, so Volume of slice π r 2 Δ h cm 3. We express r in terms of h using the Pythagorean Theorem as in Example 2. From Figure 8.10, we have so h 2 +r 2 = 7 2, Thus, Volume of slice π r 2 Δ h = π (7 2 h 2 )Δ h cm 3. Page 5 of 17

Summing the volumes of all slices gives: Volume π r 2 Δ h = π (7 2 h 2 )Δ h cm 3. As the thickness of each slice tends to zero, the sum becomes a definite integral. Since the radius of the hemisphere is 7, we know that h varies from 0 to 7, so these are the limits of integration: Notice that the volume of the hemisphere is half of, as we expected. Figure 8.10 Vertical cut through center of hemisphere showing relation between r i and h i We now use slicing to find the volume of a pyramid. We do not use the formula,, for the volume of a pyramid of height h and square base of side length b, but our approach can be used to derive that formula. Example 5 Solution Compute the volume, in cubic feet, of the Great Pyramid of Egypt, whose base is a square 755 feet by 755 feet and whose height is 410 feet. We slice the pyramid horizontally. creating square slices with thickness Δ h. The bottom layer is a square slice 755 feet by 755 feet and volume about (755) 2 Δ h ft 3. As we move up the pyramid, the layers have shorter side lengths. We divide the height into n subintervals of length Δ h. See Figure 8.11. Let s be the side length of the slice at height h ; then Volume of slice s 2 Δ h ft 3. Figure 8.11 The Great Pyramid Page 6 of 17

Interactive Exploration: Forming a Pyramid with Square Cross-sections We express s as a function of h using the vertical cross-section in Figure 8.12. By similar triangles, we get Thus, and the total volume, V, is approximated by adding the volumes of the n layers: Figure 8.12 Cross-section relating s and h As the thickness of each slice tends to zero, the sum becomes a definite integral. Finally, since h varies from 0 to 410, the height of the pyramid, we have Note that, as expected. Exercises and Problems for Section 8.1 Exercises 1. (a) Write a Riemann sum approximating the area of the region in Figure 8.13, using vertical strips as shown. Page 7 of 17

Figure 8.13 (b) Evaluate the corresponding integral. 2. (a) Write a Riemann sum approximating the area of the region in Figure 8.14, using vertical strips as shown. Figure 8.14 (b) Evaluate the corresponding integral. 3. (a) Write a Riemann sum approximating the area of the region in Figure 8.15, using horizontal strips as shown. Figure 8.15 (b) Evaluate the corresponding integral. Page 8 of 17

4. (a) Write a Riemann sum approximating the area of the region in Figure 8.16, using horizontal strips as shown. (b) 5. Figure 8.16 Evaluate the corresponding integral. In Exercises 5-12, write a Riemann sum and then a definite integral representing the area of the region, using the strip shown. Evaluate the integral exactly. 6. Page 9 of 17

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10. 11. 12. In Exercises 13-18, write a Riemann sum and then a definite integral representing the volume of the region, using the slice shown. Evaluate the integral exactly. (Regions are parts of cones, cylinders, spheres, and pyramids.) http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 11 of 17

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16. 17. 18. Problems The integrals in Problems 19-22 represent the area of either a triangle or part of a circle, and the variable of integration measures a distance. In each case, say which shape is represented, and give the radius of the circle or the base and height of the triangle. Make a sketch to support your answer showing the variable and all other relevant quantities. http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 13 of 17

19. 20. 21. 22. 23. The integral represents the area of a region between two curves in the plane. Make a sketch of this region. In Problems 24-27, construct and evaluate definite integral(s) representing the area of the region described, using: (a) Vertical slices (b) Horizontal slices 24. 25. Enclosed by y = x 2 and y = 3x. Enclosed by y = 2x and y = 12 x and the y -axis. 26. 27. Enclosed by y = x 2 and y = 6 x and the x -axis. Enclosed by y = 2x and x = 5 and y = 6 and the x -axis. The integrals in Problems 28-31 represent the volume of either a hemisphere or a cone, and the variable of integration measures a length. In each case, say which shape is represented, and give the radius of the hemisphere or the radius and height of the cone. Make a sketch to support your answer showing the variable and all other relevant quantities. 28. 29. http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 14 of 17

30. 31. 32. 33. Find the volume of a sphere of radius r by slicing. Set up and evaluate an integral to find the volume of a cone of height 12 m and base radius 3 m. 34. Find, by slicing, a formula for the volume of a cone of height h and base radius r. 35. Figure 8.17 shows a solid with both rectangular and triangular cross sections. (a) Slice the solid parallel to the triangular faces. Sketch one slice and calculate its volume in terms of x, the distance of the slice from one end. Then write and evaluate an integral giving the volume of the solid. (b) Repeat part a for horizontal slices. Instead of x, use h, the distance of a slice from the top. Figure 8.17 36. A rectangular lake is 150 km long and 3 km wide. The vertical cross-section through the lake in Figure 8.18 shows that the lake is 0.2 km deep at the center. (These are the approximate dimensions of Lake Mead, the largest reservoir in the US, which provides water to California, Nevada, and Arizona.) Set up and evaluate a definite integral giving the total volume of water in the lake. http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 15 of 17

Figure 8.18 Not to scale 37. A dam has a rectangular base 1400 meters long and 160 meters wide. Its cross-section is shown in Figure 8.19. (The Grand Coulee Dam in Washington state is roughly this size.) By slicing horizontally, set up and evaluate a definite integral giving the volume of material used to build this dam. Figure 8.19 Not to scale Strengthen Your Understanding In Problems 38-39, explain what is wrong with the statement. 38. To find the area between the line y = 2x, the y -axis, and the line y = 8 using horizontal slices, evaluate the integral. 39. The volume of the sphere of radius 10 centered at the origin is given by the integral. In Problems 40-41, give an example of: 40. A region in the plane where it is easier to compute the area using horizontal slices than it is with vertical slices. Sketch the region. 41. A triangular region in the plane for which both horizontal and vertical slices work just as easily. http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 16 of 17

In Problems 42-45, are the statements true or false? Give an explanation for your answer. 42. The integral represents the volume of a sphere of radius 3. 43. The integral gives the volume of a cone of radius r and height h. 44. 45. The integral gives the volume of a hemisphere of radius r. A cylinder of radius r and length l is lying on its side. Horizontal slicing tells us that the volume is given by. Copyright 2013 John Wiley & Sons, Inc. All rights reserved. http://edugen.wileyplus.com/edugen/courses/crs7160/hughes-halle 0NzA4ODg2NDNjMDgtc2VjLTAwMDEueGZvcm0.enc?course=crs7160&id=ref Page 17 of 17