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APPLICATIONS OF INTEGRATION 6. 6.2 Volumes APPLICATIONS OF INTEGRATION In this section, we will learn about: Using integration to find out the volume.

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Presentation on theme: "APPLICATIONS OF INTEGRATION 6. 6.2 Volumes APPLICATIONS OF INTEGRATION In this section, we will learn about: Using integration to find out the volume."— Presentation transcript:

1 APPLICATIONS OF INTEGRATION 6

2 6.2 Volumes APPLICATIONS OF INTEGRATION In this section, we will learn about: Using integration to find out the volume of a solid.

3 In trying to find the volume of a solid, we face the same type of problem as in finding areas. VOLUMES

4 We have an intuitive idea of what volume means. However, we must make this idea precise by using calculus to give an exact definition of volume. VOLUMES

5 We start with a simple type of solid called a cylinder or, more precisely, a right cylinder. VOLUMES

6 As illustrated, a cylinder is bounded by a plane region B 1, called the base, and a congruent region B 2 in a parallel plane.  The cylinder consists of all points on line segments perpendicular to the base and join B 1 to B 2. CYLINDERS

7 If the area of the base is A and the height of the cylinder (the distance from B 1 to B 2 ) is h, then the volume V of the cylinder is defined as: V = Ah CYLINDERS

8 In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V = πr 2 h. CYLINDERS

9 If the base is a rectangle with length l and width w, then the cylinder is a rectangular box (also called a rectangular parallelepiped) with volume V = l wh. RECTANGULAR PARALLELEPIPEDS

10 For a solid S that isn’t a cylinder, we first ‘cut’ S into pieces and approximate each piece by a cylinder.  We estimate the volume of S by adding the volumes of the cylinders.  We arrive at the exact volume of S through a limiting process in which the number of pieces becomes large. IRREGULAR SOLIDS

11 We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S. IRREGULAR SOLIDS

12 Let A(x) be the area of the cross-section of S in a plane P x perpendicular to the x-axis and passing through the point x, where a ≤ x ≤ b.  Think of slicing S with a knife through x and computing the area of this slice. IRREGULAR SOLIDS

13 The cross-sectional area A(x) will vary as x increases from a to b. IRREGULAR SOLIDS

14 We divide S into n ‘slabs’ of equal width ∆x using the planes P x1, P x2,... to slice the solid.  Think of slicing a loaf of bread. IRREGULAR SOLIDS

15 If we choose sample points x i * in [x i - 1, x i ], we can approximate the i th slab S i (the part of S that lies between the planes and ) by a cylinder with base area A(x i *) and ‘height’ ∆x. IRREGULAR SOLIDS

16 The volume of this cylinder is A(x i *). So, an approximation to our intuitive conception of the volume of the i th slab S i is: IRREGULAR SOLIDS

17 Adding the volumes of these slabs, we get an approximation to the total volume (that is, what we think of intuitively as the volume):  This approximation appears to become better and better as n → ∞.  Think of the slices as becoming thinner and thinner. IRREGULAR SOLIDS

18 Therefore, we define the volume as the limit of these sums as n → ∞). However, we recognize the limit of Riemann sums as a definite integral and so we have the following definition. IRREGULAR SOLIDS

19 Let S be a solid that lies between x = a and x = b. If the cross-sectional area of S in the plane P x, through x and perpendicular to the x-axis, is A(x), where A is a continuous function, then the volume of S is: DEFINITION OF VOLUME

20 When we use the volume formula, it is important to remember that A(x) is the area of a moving cross-section obtained by slicing through x perpendicular to the x-axis. VOLUMES

21 Notice that, for a cylinder, the cross-sectional area is constant: A(x) = A for all x.  So, our definition of volume gives:  This agrees with the formula V = Ah. VOLUMES

22 Show that the volume of a sphere of radius r is Example 1 SPHERES

23 If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius, from the Pythagorean Theorem, is: Example 1 SPHERES

24 So, the cross-sectional area is: Example 1 SPHERES

25 Using the definition of volume with a = -r and b = r, we have: (The integrand is even.) Example 1 SPHERES

26 The figure illustrates the definition of volume when the solid is a sphere with radius r = 1.  From the example, we know that the volume of the sphere is  The slabs are circular cylinders, or disks. SPHERES

27 The three parts show the geometric interpretations of the Riemann sums when n = 5, 10, and 20 if we choose the sample points x i * to be the midpoints. SPHERES

28 Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume. SPHERES

29 Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 1. Illustrate the definition of volume by sketching a typical approximating cylinder. Example 2 VOLUMES

30 The region is shown in the first figure. If we rotate about the x-axis, we get the solid shown in the next figure.  When we slice through the point x, we get a disk with radius. VOLUMES Example 2

31 The area of the cross-section is: The volume of the approximating cylinder (a disk with thickness ∆x) is: Example 2 VOLUMES

32 The solid lies between x = 0 and x = 1. So, its volume is: VOLUMES Example 2

33 Find the volume of the solid obtained by rotating the region bounded by y = x 3, y = 8, and x = 0 about the y-axis. Example 3 VOLUMES

34 As the region is rotated about the y-axis, it makes sense to slice the solid perpendicular to the y-axis and thus to integrate with respect to y.  Slicing at height y, we get a circular disk with radius x, where VOLUMES Example 3

35 So, the area of a cross-section through y is: The volume of the approximating cylinder is: Example 3 VOLUMES

36 Since the solid lies between y = 0 and y = 8, its volume is: Example 3 VOLUMES

37 The region R enclosed by the curves y = x and y = x 2 is rotated about the x-axis. Find the volume of the resulting solid. Example 4 VOLUMES

38 The curves y = x and y = x 2 intersect at the points (0, 0) and (1, 1).  The region between them, the solid of rotation, and cross-section perpendicular to the x-axis are shown. VOLUMES Example 4

39 A cross-section in the plane P x has the shape of a washer (an annular ring) with inner radius x 2 and outer radius x. Example 4 VOLUMES

40 Thus, we find the cross-sectional area by subtracting the area of the inner circle from the area of the outer circle: VOLUMES Example 4

41 Thus, we have: Example 4 VOLUMES

42 Find the volume of the solid obtained by rotating the region in Example 4 about the line y = 2. Example 5 VOLUMES

43 Again, the cross-section is a washer. This time, though, the inner radius is 2 – x and the outer radius is 2 – x 2. VOLUMES Example 5

44 The cross-sectional area is: Example 5 VOLUMES

45 So, the volume is: VOLUMES Example 5

46 The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a region about a line. SOLIDS OF REVOLUTION

47 In general, we calculate the volume of a solid of revolution by using the basic defining formula SOLIDS OF REVOLUTION

48 We find the cross-sectional area A(x) or A(y) in one of the following two ways. SOLIDS OF REVOLUTION

49 If the cross-section is a disk, we find the radius of the disk (in terms of x or y) and use: A = π(radius) 2 WAY 1

50 If the cross-section is a washer, we first find the inner radius r in and outer radius r out from a sketch.  Then, we subtract the area of the inner disk from the area of the outer disk to obtain: A = π(outer radius) 2 – π(outer radius) 2 WAY 2

51 Find the volume of the solid obtained by rotating the region in Example 4 about the line x = -1. Example 6 SOLIDS OF REVOLUTION

52 The figure shows the horizontal cross-section. It is a washer with inner radius 1 + y and outer radius Example 6 SOLIDS OF REVOLUTION

53 So, the cross-sectional area is: Example 6 SOLIDS OF REVOLUTION

54 The volume is: Example 6 SOLIDS OF REVOLUTION

55 In the following examples, we find the volumes of three solids that are not solids of revolution. VOLUMES

56 The figure shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid. Example 7 VOLUMES

57 Let’s take the circle to be x 2 + y 2 = 1. The solid, its base, and a typical cross-section at a distance x from the origin are shown. Example 7 VOLUMES

58 As B lies on the circle, we have So, the base of the triangle ABC is |AB| = Example 7 VOLUMES

59 Since the triangle is equilateral, we see that its height is VOLUMES Example 7

60 Thus, the cross-sectional area is : VOLUMES Example 7

61 The volume of the solid is: Example 7 VOLUMES

62 Find the volume of a pyramid whose base is a square with side L and whose height is h. Example 8 VOLUMES

63 We place the origin O at the vertex of the pyramid and the x-axis along its central axis.  Any plane P x that passes through x and is perpendicular to the x-axis intersects the pyramid in a square with side of length s. VOLUMES Example 8

64 We can express s in terms of x by observing from the similar triangles that Therefore, s = Lx/h  Another method is to observe that the line OP has slope L/(2h)  So, its equation is y = Lx/(2h) Example 8 VOLUMES

65 Thus, the cross-sectional area is: VOLUMES Example 8

66 The pyramid lies between x = 0 and x = h. So, its volume is: Example 8 VOLUMES

67 In the example, we didn’t need to place the vertex of the pyramid at the origin.  We did so merely to make the equations simple. NOTE

68 Instead, if we had placed the center of the base at the origin and the vertex on the positive y-axis, as in the figure, you can verify that we would have obtained the integral: NOTE

69 A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of 30° along a diameter of the cylinder. Find the volume of the wedge. Example 9 VOLUMES

70 If we place the x-axis along the diameter where the planes meet, then the base of the solid is a semicircle with equation -4 ≤ x ≤ 4 VOLUMES Example 9

71 A cross-section perpendicular to the x-axis at a distance x from the origin is a triangle ABC, whose base is and whose height is |BC| = y tan 30 ° = Example 9 VOLUMES

72 Thus, the cross-sectional area is: VOLUMES Example 9

73 The volume is: Example 9 VOLUMES


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