Copyright © Cengage Learning. All rights reserved. 8.2 Area of a Surface of Revolution.

Slides:



Advertisements
Similar presentations
Volumes by Slicing: Disks and Washers
Advertisements

Copyright © Cengage Learning. All rights reserved.
Applications of Integration 6. Volumes Volumes In trying to find the volume of a solid we face the same type of problem as in finding areas. We.
 A k = area of k th rectangle,  f(c k ) – g(c k ) = height,  x k = width. 6.1 Area between two curves.
Applications of Integration Copyright © Cengage Learning. All rights reserved.
7.1 Areas Between Curves To find the area: divide the area into n strips of equal width approximate the ith strip by a rectangle with base Δx and height.
APPLICATIONS OF INTEGRATION Volumes by Cylindrical Shells APPLICATIONS OF INTEGRATION In this section, we will learn: How to apply the method of.
The Shell Method Volumes by Cylindrical Shells By Christine Li, Per. 4.
APPLICATIONS OF INTEGRATION
Applications of Integration
Volume: The Disk Method
10 Applications of Definite Integrals Case Study
FURTHER APPLICATIONS OF INTEGRATION
7.3 Volumes by Cylindrical Shells
7.2 Volumes APPLICATIONS OF INTEGRATION In this section, we will learn about: Using integration to find out the volume of a solid.
Applications of Integration Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved. 10 Parametric Equations and Polar Coordinates.
Conics, Parametric Equations, and Polar Coordinates Copyright © Cengage Learning. All rights reserved.
10 Conics, Parametric Equations, and Polar Coordinates
6.3 Volumes by Cylindrical Shells APPLICATIONS OF INTEGRATION In this section, we will learn: How to apply the method of cylindrical shells to find out.
Copyright © Cengage Learning. All rights reserved. 15 Multiple Integrals.
DOUBLE INTEGRALS IN POLAR COORDINATES
Conics, Parametric Equations, and Polar Coordinates Copyright © Cengage Learning. All rights reserved.
Section 6.5 Area of a Surface of Revolution. All graphics are attributed to:  Calculus,10/E by Howard Anton, Irl Bivens, and Stephen Davis Copyright.
Copyright © Cengage Learning. All rights reserved. 8 Further Applications of Integration.
Copyright © Cengage Learning. All rights reserved The Area Problem.
Copyright © Cengage Learning. All rights reserved. 15 Multiple Integrals.
Arc Length and Surfaces of Revolution
7.4 Length of a Plane Curve y=f(x) is a smooth curve on [a, b] if f ’ is continuous on [a, b].
Lesson 9-1: Area of 2-D Shapes 1 Part 1 Area of 2-D Shapes.
Volume: The Disk Method
Solids of Revolution Disk Method
Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, Disk and Washer Methods Limerick Nuclear Generating Station, Pottstown,
Conics, Parametric Equations, and Polar Coordinates 10 Copyright © Cengage Learning. All rights reserved.
Applications of Integration Copyright © Cengage Learning. All rights reserved.
8.2 Area of a Surface of Revolution
Copyright © Cengage Learning. All rights reserved. 16 Vector Calculus.
6.3 – Volumes of Cylindrical Shells. Derivation Assume you have a functions similar to the one shown below and assume the f is to difficult to solve for.
Tangents.
Volume: The Shell Method
Section 9.2 Area of a Surface of Revolution. THE AREA OF A FRUSTUM The area of the frustum of a cone is given by.
8.1 Arc Length and Surface Area Thurs Feb 4 Do Now Find the volume of the solid created by revolving the region bounded by the x-axis, y-axis, and y =
Chapter Area between Two Curves 7.2 Volumes by Slicing; Disks and Washers 7.3 Volumes by Cylindrical Shells 7.4 Length of a Plane Curve 7.5 Area.
15 Copyright © Cengage Learning. All rights reserved. Vector Analysis.
Applications of Integration 6. More About Areas 6.1.
Copyright © Cengage Learning. All rights reserved. 5.2 Volumes
Conics, Parametric Equations, and Polar Coordinates 10 Copyright © Cengage Learning. All rights reserved.
Prism & Pyramids. Lesson 9-2: Prisms & Pyramids2 Right Prism Lateral Area of a Right Prism (LA) = ph Surface Area (SA) = ph + 2B = [Lateral Area + 2 (area.
7.4 Day 2 Surface Area Greg Kelly, Hanford High School, Richland, Washington(Photo not taken by Vickie Kelly)
7 Applications of Integration
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
In this section, we will learn about: Using integration to find out
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
10 Conics, Parametric Equations, and Polar Coordinates
7 Applications of Integration
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Conics, Parametric Equations, and Polar Coordinates
Copyright © Cengage Learning. All rights reserved.
Copyright © Cengage Learning. All rights reserved.
Applications of Integration
Presentation transcript:

Copyright © Cengage Learning. All rights reserved. 8.2 Area of a Surface of Revolution

2 A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution. We want to define the area of a surface of revolution. If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A.

3 Area of a Surface of Revolution For Example: The lateral surface area of a circular cylinder with radius r and height h is taken to be A = 2  rh, because we can imagine cutting the cylinder and unrolling it to obtain a rectangle with dimensions 2  r and h. Figure 1

4 Area of a Surface of Revolution Likewise, we can take a circular cone with base radius r and slant height l, cut it along the dashed line and flatten it to form a sector of a circle with radius l and central angle  = 2  r  l. The area of a sector of a circle A = l 2  = =  r l Figure 2 Therefore we define the lateral surface area of a cone to be A =  r l.

5 Area of a Surface of Revolution What about more complicated surfaces of revolution? If we follow the strategy we used with arc length, we can approximate the original curve by a polygon.

6 Area of a Surface of Revolution When this polygon is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. By taking a limit, we can determine the exact surface area. The approximating surface consists of a number of bands, each formed by rotating a line segment about an axis.

7 Area of a Surface of Revolution To find the surface area, each of these bands can be considered a portion of a circular cone, as shown in Figure 3. Figure 3

8 Area of a Surface of Revolution The area of the band (or frustum of a cone) with slant height l and upper and lower radii r 1 and r 2 is found by subtracting the areas of two cones: A =  r 2 (l 1 + l ) –  r 1 l 1 =  [(r 2 – r 1 ) l 1 + r 2 l ] From similar triangles we have which gives r 2 l 1 = r 1 l 1 + r 1 l or (r 2 – r 1 )l 1 = r 1 l

9 Area of a Surface of Revolution Putting this in Equation 1, we get A =  (r 1 l + r 2 l ) or where r = (r 1 + r 2 ) is the average radius of the band.

10 Area of a Surface of Revolution Now we apply this formula to our strategy. Consider the surface shown in Figure 4, which is obtained by rotating the curve y = f (x), a  x  b, about the x-axis, where f is positive and has a continuous derivative. (a) Surface of revolution (b) Approximating band Figure 4

11 Area of a Surface of Revolution In order to define its surface area, we divide the interval [a, b] into n subintervals with endpoints x 0, x 1,..., x n and equal width  x. If y i = f (x i ), then the point P i (x i, y i ) lies on the curve. The part of the surface between x i – 1 and x i is approximated by taking the line segment P i – 1 P i and rotating it about the x-axis.

12 Area of a Surface of Revolution The result is a band with slant height l = |P i – 1 P i | and average radius r = (y i – 1 + y i ) so, by Formula 2, its surface area is We have where x i  is some number in [x i – 1, x i ].

13 Area of a Surface of Revolution When  x is small, we have y i = f (x i )  f (x i  ) and also Y i – 1 = f (x i – 1 )  f (x i  ), since f is continuous. Therefore and so an approximation to what we think of as the area of the complete surface of revolution is

14 Area of a Surface of Revolution This approximation appears to become better as n  and, recognizing as a Riemann sum for the function g(x) = 2  f (x), we have Therefore, in the case where f is positive and has a continuous derivative, we define the surface area of the surface obtained by rotating the curve y = f (x), a  x  b, about the x-axis as

15 Area of a Surface of Revolution With the Leibniz notation for derivatives, this formula becomes If the curve is described as x = g(y), c  y  d, then the formula for surface area becomes

16 Area of a Surface of Revolution These formulas can be remembered by thinking of 2  y or 2  x as the circumference of a circle traced out by the point (x, y) on the curve as it is rotated about the x-axis or y-axis, respectively. (a) Rotation about x-axis: S = ∫ 2  y ds (b) Rotation about y-axis: S = ∫ 2  x ds Figure 5

17 Example 1 The curve y =, –1  x  1, is an arc of the circle x 2 + y 2 = 4. Find the area of the surface obtained by rotating this arc about the x-axis. (The surface is a portion of a sphere of radius 2. See Figure 6.) Figure 6

18 Example 1 – Solution We have and so, by Formula 5, the surface area is

19 Example 1 – Solution = 4  1 dx = 4  (2) = 8  cont’d

20 Example 3 Find the area of the surface generated by rotating the curve y = e x, 0  x  1, about the x-axis. Solution: Using Formula 5 with y = e x and = e x we have

21 Example 3 – Solution =  [sec  tan  + ln(sec  + tan  ) – – ln( + 1)] cont’d (where u = e x ) (where u = tan  and  = tan –1 e)

22 Example 3 – Solution Since tan  = e, we have sec 2  = 1 + tan 2  = 1 + e 2 and S =  [ + ln( ) – – ln( + 1)] cont’d

23 Video Examples: Revolving around the y-axis: Revolving around the x-axis: