Lecture 21. Section 6.1 More on Area Section 6.2 Volume by Parallel Cross Section. Jiwen He. Department of Mathematics, University of Houston
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1 Section 6.1 Section 6.2 Lecture 21 Section 6.1 More on Area Section 6.2 Volume by Parallel Cross Section Jiwen He Department of Mathematics, University of Houston math.uh.edu/ jiwenhe/math1431 November 18, /
2 Test 3 Section 6.1 Section 6.2 Test 3: Dec. 4-6 in CASA November 18, /
3 Final Exam Section 6.1 Section 6.2 Final Exam: Dec in CASA November 18, /
4 Section 6.1 Section 6.2 Review for Test 3 Review for Test 3 by the College Success Program. Friday, November 21 2:30 3:30pm in the basement of the library by the C-site. November 18, /
5 Online Quizzes Section 6.1 Section 6.2 Online Quizzes are available on CourseWare. November 18, /
6 Quiz 1 Section 6.1 Section 6.2 What is today? a. Monday b. Wednesday c. Friday d. None of these November 18, /
![Section 6.1 Section 6.2 The Average value of f Let f avg denote the average or mean value of f on [a, b]. Then f avg = 1 b a b a f (x) dx.](/docs-images/90/102011292/images/7-1.jpg "The First Mean-Value Theorems for Integrals If f is continous on [a, b], then there is at least one number c in (a, b) for which f (c) = f avg.")
7 Section 6.1 Section 6.2 The Average value of f Let f avg denote the average or mean value of f on [a, b]. Then f avg = 1 b a b a f (x) dx. The First Mean-Value Theorems for Integrals If f is continous on [a, b], then there is at least one number c in (a, b) for which f (c) = f avg. November 18, /
8 Quiz 2 Section 6.1 Section 6.2 The graph of y = f (x) is shown below, where Ω 1 has area 2, Ω 2 has area 3, Ω 3 has area 8, and Ω 4 has area 4. Give the average value of f on the interval [d, b] with d = 3 and b = 6. a. 1/3 b. 2/3 c. 1 d. 4/3 e. None of these November 18, /
9 Quiz 3 Section 6.1 Section 6.2 The graph of y = f (x) is shown below, where Ω 1 has area 2, Ω 2 has area 3, Ω 3 has area 8, and Ω 4 has area 4. How many values of c satisfy the conclusion of the mean value theorem for integrals on the interval [d, b] with d = 3 and b = 6. a. 1 b. 2 c. 3 d. None of these November 18, /
10 Quiz 4 Section 6.1 Section 6.2 The graph of y = f (x) is shown below, where Ω 1 has area 2, Ω 2 has area 3, Ω 3 has area 8, and Ω 4 has area 4. Give the average value of f on the interval [a, b] with a = 1 and b = 6. a. 1/5 b. 2/5 c. 3/5 d. 4/5 e. None of these November 18, /
11 Quiz 5 Section 6.1 Section 6.2 The graph of y = f (x) is shown below, where Ω 1 has area 2, Ω 2 has area 3, Ω 3 has area 8, and Ω 4 has area 4. How many values of c satisfy the conclusion of the mean value theorem for integrals on the interval [a, b] with d = 1 and b = 6. a. 1 b. 2 c. 3 d. None of these November 18, /
12 Section 6.1 Section 6.2 Representative Rectangle, Riemann Sum and Area: f 0 Rectangle: f (x i ) x i Riemann Sum: f (x i ) x i b a f (x) dx = lim f (x i ) x i. P 0 area = b a f (x) dx f (x i ) x i. November 18, /
![Rectangle Area [f (x i ) g(x i )] x i Riemann Sum [f (x i](/docs-images/90/102011292/images/13-2.jpg ") g(x i )] x i area(ω) = b a [f (x) g(x)] dx = lim [f (x")
13 Section 6.1 Section 6.2 Area by Integration with Respect to x: f (x) g(x) Rectangle Area [f (x i ) g(x i )] x i Riemann Sum [f (x i ) g(x i )] x i area(ω) = b a [f (x) g(x)] dx = lim [f (x i ) g(xi )] x i. P 0 November 18, /
14 Section 6.1 Section 6.2 Area by Integration with Respect to y: F (y) G(y) Rectangle Area [F (yi ) G(y i )] y i Riemann Sum [F (y i ) G(yi )] y i area(ω) = d c [F (y) G(y)] dy = lim [F (y i ) G(yi )] y i. P 0 November 18, /
15 Example Section 6.1 Section 6.2 Example Find the area of the shaded region shown in the figure below. November 18, /
16 Example Section 6.1 Section 6.2 Example Find the area of the shaded region shown in the figure below. November 18, /
17 Example Example Section 6.1 Section 6.2 Find the area of the shaded region shown in the figure below by integrating with respect to x. November 18, /
18 Example Example Section 6.1 Section 6.2 Find the area of the shaded region shown in the figure below by integrating with respect to y. November 18, /
19 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Right Cylinder with Cross Section Volume of a Right Cylinder with Cross Section V = A h = (cross-sectional area) height November 18, /
20 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Right Circular Cylinder and Rectangular Box Volume of a Right Circular Cylinder V = πr 2 h = (cross-sectional area) height Volume of a Rectangular Box V = l w h = (cross-sectional area) height November 18, /
21 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Volume by Parallel Cross Section Cylinder Volume A(x i ) x i Riemann Sum A(x i ) x i V = b a A(x) dx = lim A(x i ) x i P 0 November 18, /
22 Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Example Find the volume of the pyramid shown in the figure below. November 18, /
23 Example Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer The base of a solid is the region bounded by the ellipse x 2 a 2 + y 2 b 2 = 1. Find the volume of the solid given that each cross section is an isosceles triangle with base in the region and altitude equal to one-half the base. November 18, /
24 Example Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer The base of a solid is the region between the parabolas x = y 2 and x = 3 2y 2. Find the volume of the solid given that the cross sections are squares. November 18, /
25 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Solid of Revolution About the x-axis Cylinder Volume: π[f (x i )]2 x i Riemann Sum: π[f (x i )]2 x i V = b a π[f (x)] 2 dx = lim π[f (x i )] 2 x i. P 0 November 18, /
26 Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Example Find the volume of the cone shown in the figure below. November 18, /
27 Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Example Find the volume of a sphere of radius r by revolving about the x-axis the region below the graph of f (x) = r 2 x 2, r x r. November 18, /
28 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Solid of Revolution About the y-axis Cylinder Volume: π[g(y i )]2 y i Riemann Sum: π[g(y i )]2 y i V = d c π[g(y)] 2 dy = lim π[g(y i )] 2 y i. P 0 November 18, /
29 Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Example Find the volume of the solid shown in the figure below. November 18, /
30 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Solid of Revolution About the x-axis Cylinder Volume: π([f (x i )]2 [g(x i )]2 ) x i Riemann Sum: π([f (x i )]2 [g(x i )]2 ) x i V = b a π([f (x)] 2 [g(x)] 2 ) dx = lim π([f (x i )] 2 [g(xi )] 2 ) x i P 0 November 18, /
31 Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Solid of Revolution About the y-axis Cylinder Volume: π([f (yi )]2 [G(yi )]2 ) y i Riemann Sum: π([f (yi )]2 [G(yi )]2 ) y i V = d c π([f (y)] 2 [G(y)] 2 ) dy = lim π([f (y i )] 2 [G(yi )] 2 ) y i P 0 November 18, /
32 Example Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Find the volume of the solid generated by revolving the region between y = x 2 and y = 2x about the x-axis. November 18, /
33 Example Example Section 6.1 Section 6.2 Volume-Cross Section Revolution-Disk Revolution-Washer Find the volume of the solid generated by revolving the region between y = x 2 and y = 2x about the y-axis. November 18, 2008 /
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