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Applications of Integration In this chapter we explore some of the applications of the definite integral by using it for 1.Computing the area between curves.

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Presentation on theme: "Applications of Integration In this chapter we explore some of the applications of the definite integral by using it for 1.Computing the area between curves."— Presentation transcript:

1 Applications of Integration In this chapter we explore some of the applications of the definite integral by using it for 1.Computing the area between curves 2.Computing the volumes of solids 3.Computing the work done by a varying force 4.Computing average value of a function The common theme is the following general method, which is similar to the one we used to find areas under curves: We break up a Q quantity into a large number of small parts. We next approximate each small part by a quantity of the form and thus approximate Q by a Riemann sum. Then we take the limit and express Q as an integral Finally we evaluate the integral using the Fundamental Theorem of Calculus or the Midpoint Rule.

2 Area Between Two Curves asses/Riemann/Diff.htmlhttp://www.math.tamu.edu/AppliedCalc/Cl asses/Riemann/Diff.html

3 Example

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6 How can we find the area between these two curves? We could split the area into several sections, use subtraction and figure it out, but there is an easier way.

7 Consider a very thin vertical strip. The length of the strip is: or Since the width of the strip is a very small change in x, we could call it dx.

8 Since the strip is a long thin rectangle, the area of the strip is: If we add all the strips, we get:

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10 The formula for the area between curves is: We will use this so much, that you won’t need to “memorize” the formula!

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13 Some regions are best treated by regarding x as a function of y

14 If we try vertical strips, we have to integrate in two parts: We can find the same area using a horizontal strip. Since the width of the strip is dy, we find the length of the strip by solving for x in terms of y.

15 We can find the same area using a horizontal strip. Since the width of the strip is dy, we find the length of the strip by solving for x in terms of y. length of strip width of strip

16 General Strategy for Area Between Curves: 1 Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.) Sketch the curves. 2 3 Write an expression for the area of the strip. (If the width is dx, the length must be in terms of x. If the width is dy, the length must be in terms of y. 4 Find the limits of integration. (If using dx, the limits are x values; if using dy, the limits are y values.) 5 Integrate to find area. 

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18 3 3 3 Find the volume of the pyramid: Consider a horizontal slice through the pyramid. s dh The volume of the slice is s 2 dh. If we put zero at the top of the pyramid and make down the positive direction, then s=h. 0 3 h This correlates with the formula:

19 Method of Slicing: 1 Find a formula for V ( x ). (Note that I used V ( x ) instead of A(x).) Sketch the solid and a typical cross section. 2 3 Find the limits of integration. 4 Integrate V ( x ) to find volume.

20 x y A 45 o wedge is cut from a cylinder of radius 3 as shown. Find the volume of the wedge. You could slice this wedge shape several ways, but the simplest cross section is a rectangle. If we let h equal the height of the slice then the volume of the slice is: Since the wedge is cut at a 45 o angle: x h 45 o Since

21 x y Even though we started with a cylinder,  does not enter the calculation!

22 Cavalieri’s Theorem: Two solids with equal altitudes and identical parallel cross sections have the same volume. Identical Cross Sections 

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27 Suppose I start with this curve. My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape. So I put a piece of wood in a lathe and turn it to a shape to match the curve.

28 How could we find the volume of the cone? One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes. The volume of each flat cylinder (disk) is: In this case: r= the y value of the function thickness = a small change in x = dx

29 The volume of each flat cylinder (disk) is: If we add the volumes, we get:

30 This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk. If the shape is rotated about the x-axis, then the formula is: Since we will be using the disk method to rotate shapes about other lines besides the x-axis, we will not have this formula on the formula quizzes. A shape rotated about the y-axis would be:

31 31 Revolving a Function Consider a function f(x) on the interval [a, b] Now consider revolving that segment of curve about the x axis What kind of functions generated these solids of revolution? f(x) a b

32 32 Disks We seek ways of using integrals to determine the volume of these solids Consider a disk which is a slice of the solid –What is the radius –What is the thickness –What then, is its volume? dx f(x)

33 33 Disks To find the volume of the whole solid we sum the volumes of the disks Shown as a definite integral f(x) a b

34 The region between the curve, and the y -axis is revolved about the y -axis. Find the volume. y x We use a horizontal disk. The thickness is dy. The radius is the x value of the function. volume of disk

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36 The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis: The volume can be calculated using the disk method with a horizontal disk.

37 The region bounded by and is revolved about the y-axis. Find the volume. The “disk” now has a hole in it, making it a “washer”. If we use a horizontal slice: The volume of the washer is: outer radius inner radius

38 This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle. The washer method formula is: Like the disk method, this formula will not be on the formula quizzes. I want you to understand the formula.

39 If the same region is rotated about the line x = 2 : The outer radius is: R The inner radius is: r 

40 Find the volume of the region bounded by,, and revolved about the y - axis. We can use the washer method if we split it into two parts: outer radius inner radius thickness of slice cylinder Japanese Spider Crab Georgia Aquarium, Atlanta

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47 The solids in Examples 1–5 are all called solids of revolution because they are obtained by revolving a region about a line. In general, we calculate the volume of a solid of revolution by using the basic defining formula We find the cross-sectional area in one of the following ways:

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49 If we take a vertical sliceand revolve it about the y-axis we get a cylinder. cross section If we add all of the cylinders together, we can reconstruct the original object. Here is another way we could approach this problem:

50 cross section The volume of a thin, hollow cylinder is given by: r is the x value of the function. h is the y value of the function. thickness is dx.

51 cross section If we add all the cylinders from the smallest to the largest: This is called the shell method because we use cylindrical shells.

52 Find the volume generated when this shape is revolved about the y axis. We can’t solve for x, so we can’t use a horizontal slice directly.

53 Shell method: If we take a vertical slice and revolve it about the y-axis we get a cylinder.

54 Note:When entering this into the calculator, be sure to enter the multiplication symbol before the parenthesis.

55 55 Shell Method Based on finding volume of cylindrical shells –Add these volumes to get the total volume Dimensions of the shell –Radius of the shell –Thickness of the shell –Height

56 56 The Shell Consider the shell as one of many of a solid of revolution The volume of the solid made of the sum of the shells f(x) g(x) x f(x) – g(x) dx

57 57 Try It Out! Consider the region bounded by x = 0, y = 0, and

58 58 Hints for Shell Method Sketch the graph over the limits of integration Draw a typical shell parallel to the axis of revolution Determine radius, height, thickness of shell Volume of typical shell Use integration formula

59 59 Rotation About x-Axis Rotate the region bounded by y = 4x and y = x 2 about the x-axis What are the dimensions needed? –radius –height –thickness radius = y thickness = dy

60 60 Rotation About Noncoordinate Axis Possible to rotate a region around any line Rely on the basic concept behind the shell method x = a f(x) g(x)

61 61 Rotation About Noncoordinate Axis What is the radius? What is the height? What are the limits? The integral: x = a f(x) g(x) a – x f(x) – g(x) x = c r c < x < a

62 62 Try It Out Rotate the region bounded by 4 – x 2, x = 0 and, y = 0 about the line x = 2 Determine radius, height, limits 4 – x 2 r = 2 - x

63 63 Try It Out Integral for the volume is

64 Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer 1. Review Problems

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68 Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line

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70 Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the Specific line. Sketch the region and a typical shell

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74 Average value of a function The average value of function f on the interval [a, b] is defined as Note: For a positive function, we can think of this definition as saying area/width = average height Example: Find the average value of f(x)=x 3 on [0,2].

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77 The Mean Value Theorem for Integrals Theorem: If f is continuous on [a, b], then there exists a number c in [a, b] such that Example: Find c such that f ave =f(c) for f(x)=x 3 on [0,2]. From previous slide, f(c)=f ave =2. Thus, c 3 =2, so

78 4. Computing average value of a function

79 Find the average value of the function on the given interval

80 1. 2.

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