DOCSLIB.ORG

Calculus II, Section 8.2, #4 Area of a Surface of Revolution (A) Set up An

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Calculus II, Section 8.2, #4 Area of a Surface of Revolution (A) Set up An Calculus II, Section 8.2, #4 Area of a Surface of Revolution (a) Set up an integral for the area of the surface obtained by rotating the curve about (i) the x-axis and (ii) the y-axis.1 x = ln (2y + 1) , 0 ≤ y ≤ 1 (b) Use the numerical integration capability of a calculator to evaluate the surface areas correct to four decimal places. In any calculus textbook, there are a lot of formulas associated with surface area. In the eighth edition of Calculus, Early Transcendentals by James Stewart, there are five formulas set out for finding the surface area of a solid of revolution. If we try to intuitively understand the surface area concept, we can spare ourselves the need to memorize these formulas. Much like the computation of the lateral area of a cylinder (circumference·length), the area of a surface of revolution is found by circumference · arc length where we use the formula for arc length and integrate over the length of the interval of integration. Thus we have x=b dy 2 S = 2π · radius · 1+ dx s dx Zx=a or y=d dx 2 S = 2π · radius · 1+ dy s dy Zy=c For this problem, we will be finding surface areas associated with revolving the graph of x = ln (2y + 1) , 0 ≤ y ≤ 1 about the x and y axes. Let’s do some preliminary work with the function. The graph of x = ln (2y + 1) , 0 ≤ y ≤ 1 is shown y at right. A simple computation confirms when y = 0, x = ln (2 · 0+1) = 0 and when y = 1, 1 x = ln (2 · 1 + 1) = ln (3). Also, if we solve for y we get x 1 ln(3) x = ln (2y + 1) ex =eln(2y+1) ex =2y +1 ex − 1 y = 2 Since Since ex − 1 x = ln (2y + 1) y = 2 we have 1 = (ex − 1) dx 1 2 = · 2 dy 2y +1 we have dx 2 dy 1 = = ex dy 2y +1 dx 2 With this preliminary work completed, we are ready to respond to the prompt. 1Stewart, Calculus, Early Transcendentals, p. 555, #4. Calculus II Area of a Surface of Revolution (i) The diagram at right shows the curve being re- y volved about the x-axis, along with a radius. x In terms of x, the radius is y = e −1 , the arc length 2 1 2 is 1+ 1 ex dx, and we integrate from x =0to 2 radius x = ln(3). We get q x x=ln(3) ex − 1 1 2 S = 2π 1+ ex dx ln(3) 2 s 2 Zx=0 ≈ 4.2583 In terms of y, the radius is y, the arc length is 2 2 1+ 2y+1 dy, and we integrate from y =0 to yr= 1. We get y=1 2 2 S = 2π · y · 1+ dy s 2y +1 Zy=0 ≈ 4.2583 (ii) The diagram at right shows the curve being re- y volved about the y-axis, along with a radius. In terms of y, the radius is x = ln (2y +1), the arc 1 radius 2 2 length is 1+ 2y+1 dy, and we integrate from y =0 to yr= 1. We get x y=1 2 2 ln(3) S = 2π · ln (2y + 1) · 1+ dy s 2y +1 Zy=0 ≈ 5.6053 In terms of x, the radius is x, the arc length is 1 x 2 1+ 2 e dx, and we integrate from x = 0 to x = ln(3). We get q x=ln(3) 1 2 S = 2π · x · 1+ ex dx s 2 Zx=0 ≈ 5.6053.
Load more

Recommended publications
  • Arc Length. Surface Area
    Calculus 2 Lia Vas Arc Length. Surface Area. Arc Length. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: The arc length L of f(x) for a ≤ x ≤ b can be obtained by integrating the length element ds from a to b: The length element ds on a sufficiently small interval can be approximated by the hypotenuse of a triangle with sides dx and dy: p Thus ds2 = dx2 + dy2 ) ds = dx2 + dy2 and so s s Z b Z b q Z b dy2 Z b dy2 L = ds = dx2 + dy2 = (1 + )dx2 = (1 + )dx: a a a dx2 a dx2 2 dy2 dy 0 2 Note that dx2 = dx = (y ) : So the formula for the arc length becomes Z b q L = 1 + (y0)2 dx: a Area of a surface of revolution. Suppose that y = f(x) is a continuous function with a continuous derivative on [a; b]: To compute the surface area Sx of the surface obtained by rotating f(x) about x-axis on [a; b]; we can integrate the surface area element dS which can be approximated as the product of the circumference 2πy of the circle with radius y and the height that is given by q the arc length element ds: Since ds is 1 + (y0)2dx; the formula that computes the surface area is Z b q 0 2 Sx = 2πy 1 + (y ) dx: a If y = f(x) is rotated about y-axis on [a; b]; then dS is the product of the circumference 2πx of the circle with radius x and the height that is given by the arc length element ds: Thus, the formula that computes the surface area is Z b q 0 2 Sy = 2πx 1 + (y ) dx: a Practice Problems.
    [Show full text]
  • Cones, Pyramids and Spheres
    The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 12 CONES, PYRAMIDS AND SPHERES A guide for teachers - Years 9–10 June 2011 YEARS 910 Cones, Pyramids and Spheres (Measurement and Geometry : Module 12) For teachers of Primary and Secondary Mathematics 510 Cover design, Layout design and Typesetting by Claire Ho The Improving Mathematics Education in Schools (TIMES) Project 2009‑2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations. The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. © The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE‑EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution‑NonCommercial‑NoDerivs 3.0 Unported License. 2011. http://creativecommons.org/licenses/by‑nc‑nd/3.0/ The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 12 CONES, PYRAMIDS AND SPHERES A guide for teachers - Years 9–10 June 2011 Peter Brown Michael Evans David Hunt Janine McIntosh Bill Pender Jacqui Ramagge YEARS 910 {4} A guide for teachers CONES, PYRAMIDS AND SPHERES ASSUMED KNOWLEDGE • Familiarity with calculating the areas of the standard plane figures including circles. • Familiarity with calculating the volume of a prism and a cylinder. • Familiarity with calculating the surface area of a prism. • Facility with visualizing and sketching simple three‑dimensional shapes. • Facility with using Pythagoras’ theorem. • Facility with rounding numbers to a given number of decimal places or significant figures.
    [Show full text]
  • Lateral and Surface Area of Right Prisms 1 Jorge Is Trying to Wrap a Present That Is in a Box Shaped As a Right Prism
    CHAPTER 11 You will need Lateral and Surface • a ruler A • a calculator Area of Right Prisms c GOAL Calculate lateral area and surface area of right prisms. Learn about the Math A prism is a polyhedron (solid whose faces are polygons) whose bases are congruent and parallel. When trying to identify a right prism, ask yourself if this solid could have right prism been created by placing many congruent sheets of paper on prism that has bases aligned one above the top of each other. If so, this is a right prism. Some examples other and has lateral of right prisms are shown below. faces that are rectangles Triangular prism Rectangular prism The surface area of a right prism can be calculated using the following formula: SA 5 2B 1 hP, where B is the area of the base, h is the height of the prism, and P is the perimeter of the base. The lateral area of a figure is the area of the non-base faces lateral area only. When a prism has its bases facing up and down, the area of the non-base faces of a figure lateral area is the area of the vertical faces. (For a rectangular prism, any pair of opposite faces can be bases.) The lateral area of a right prism can be calculated by multiplying the perimeter of the base by the height of the prism. This is summarized by the formula: LA 5 hP. Copyright © 2009 by Nelson Education Ltd. Reproduction permitted for classrooms 11A Lateral and Surface Area of Right Prisms 1 Jorge is trying to wrap a present that is in a box shaped as a right prism.
    [Show full text]
  • Calculating Volume of a Solid
    Calculating volume of a solid Quick Rewind: how to calculate area 1. slice the region to smaller pieces (narrow, nearly rectangle) 2. calculate the area of rectangle 3. take the limit (more and more slices) Recipe for calculating volume (motivated by area) 1. slice the solid into smaler pieces (thin cut) 2. calculate the volume of each \thin cut" 3. take the limit (more and more slices) Step 2: volume ≈ area × thickness, because it is so thin Suppose that x-axis is perpendicular to the direction of cutting, and the solid is between x = a and x = b. Z b Volume= A(x)dx, where A(x) is the area of \thin cut"at x a (\thin cut"is called a cross section, a two-dimensional region) n Z b X A(x)dx = lim A(yi ) · ∆x, (Riemann sum), where ∆x = (b − a)=n, x0 = a, n!1 a i=1 x1 = x0 + ∆x, ··· , xk+1 = xk + ∆x, ··· , xn = b, and xi−1 ≤ yi ≤ xi . Slicing method of volume Example 1: (classical ones) Z h (a) circular cylinder with radius=r, height=h: A(x)dx, A(x) = π[R(x)]2, R(x) = r 0 Z h rx (b) circular cone with radius=r, height=h: A(x)dx, A(x) = π[R(x)]2, R(x) = 0 h Z r p (c) sphere with radius r: A(x)dx, A(x) = π[R(x)]2, R(x) = r 2 − x2. −r Example 2: Find the volume of the solid obtained by rotating the region bounded by y = x − x2, y = 0, x = 0 and x = 1 about x-axis.
    [Show full text]
  • Calculus Terminology
    AP Calculus BC Calculus Terminology Absolute Convergence Asymptote Continued Sum Absolute Maximum Average Rate of Change Continuous Function Absolute Minimum Average Value of a Function Continuously Differentiable Function Absolutely Convergent Axis of Rotation Converge Acceleration Boundary Value Problem Converge Absolutely Alternating Series Bounded Function Converge Conditionally Alternating Series Remainder Bounded Sequence Convergence Tests Alternating Series Test Bounds of Integration Convergent Sequence Analytic Methods Calculus Convergent Series Annulus Cartesian Form Critical Number Antiderivative of a Function Cavalieri’s Principle Critical Point Approximation by Differentials Center of Mass Formula Critical Value Arc Length of a Curve Centroid Curly d Area below a Curve Chain Rule Curve Area between Curves Comparison Test Curve Sketching Area of an Ellipse Concave Cusp Area of a Parabolic Segment Concave Down Cylindrical Shell Method Area under a Curve Concave Up Decreasing Function Area Using Parametric Equations Conditional Convergence Definite Integral Area Using Polar Coordinates Constant Term Definite Integral Rules Degenerate Divergent Series Function Operations Del Operator e Fundamental Theorem of Calculus Deleted Neighborhood Ellipsoid GLB Derivative End Behavior Global Maximum Derivative of a Power Series Essential Discontinuity Global Minimum Derivative Rules Explicit Differentiation Golden Spiral Difference Quotient Explicit Function Graphic Methods Differentiable Exponential Decay Greatest Lower Bound Differential
    [Show full text]
  • Volume of Solids
    30B Volume Solids Volume of Solids 1 30B Volume Solids The volume of a solid right prism or cylinder is the area of the base times the height. 2 30B Volume Solids Disk Method How would we find the volume of a shape like this? Volume of one penny: Volume of a stack of pennies: 3 30B Volume Solids EX 1 Find the volume of the solid of revolution obtained by revolving the region bounded by , the x-axis and the line x=9 about the x-axis. 4 30B Volume Solids EX 2 Find the volume of the solid generated by revolving the region enclosed by and about the y-axis. 5 30B Volume Solids Washer Method How would we find the volume of a washer? 6 30B Volume Solids EX 3 Find the volume of the solid generated by revolving about the x-axis the region bounded by and . 7 30B Volume Solids EX 4 Find the volume of the solid generated by revolving about the line y = 2 the region in the first quadrant bounded by these parabolas and the y-axis. (Hint: Always measure radius from the axis of revolution.) 8 30B Volume Solids Shell Method How would we find the volume of a label we peel off a can? 9 30B Volume Solids EX 5 Find the volume of the solid generated when the region bounded by these three equations is revolved about the y-axis. 10 30B Volume Solids EX 6 Find the volume of the solid generated when the region bounded by these equations is revolved about the y-axis in two ways.
    [Show full text]
  • Revolution About the X-Axis Find the Volume of the Solid of Revolution Generated by Rotating the Regions Bounded by the Curves Given Around the X-Axis
    Revolution About the x-axis Find the volume of the solid of revolution generated by rotating the regions bounded by the curves given around the x-axis. a) y = 3x − x2; y = 0 p b) y = ax; y = 0; x = a Solutions a) y = 3x − x2; y = 0 We start by sketching the region, which is a parabolic \hump" between x = 0 and x = 3 with vertex (1:5; 2:25). We could compute the volume of the solid by using shells or disks. Because we're given the equation y = 3x − x2 for y in terms of x, it seems easiest to use disks. The radius of each disk will equal the height of the curve, 3x − x2 so a disk's volume will be π(3x − x2)2 dx. The limits of integration are the x-intercepts (crossing points) x = 0 and x = 3. The volume of the solid of revolution is: Z 3 Z 3 2 2 2 3 4 π(3x − x ) dx = π 9x − 6x + x dx 0 0 � 3 1 �3 = π 3x 3 − x 4 + x 5 2 5 0 �� 3 1 � � = π 3 · 33 − · 34 + · 35 − 0 2 5 � 3 3 � = 34π 1 − + 2 5 34π = ≈ 8π: 10 This seems like a reasonable answer; 8π is the volume of a cylinder with radius 2 and height 2. p b) y = ax; y = 0; x = a We must either sketch this region by hand or choose a value for a when using an electronic graphing tool. Either way, we find that the region has the shape of half of a parabolic \hump", opening to the right, with vertex at (0; 0).
    [Show full text]
  • Comparing Surface Area & Volume
    Name ___________________________________________________________________ Period ______ Comparing Surface Area & Volume Use surface area and volume calculations to solve the following situations. 1. Find the surface area and volume of each prism. Which two have the same surface area? Which two have the same volume? 4 in. 22 in. 10 in. C A 11 in. B 18 in. 10 in. 6 in. 10 in. 6 in. 2. Sketch and label two different rectangular prisms that have the same volume. Find the surface area of each. 3. Do the following cylinders have the same surface area, the same volume, or the same surface area and volume? r = 4 cm r = 2 cm h = 9 cm h = 24 cm 4. A cylinder has a surface area of 125.6 mm2 and a volume of 100.48 mm3. If another cylinder has a radius of 5 mm and the same volume, what would be the height? 5. A cylinder has a radius of 9 yd and a height of 3 yd. A second cylinder has a radius of 6 yd and a height of 12 yd. Which statement is true? a. The cylinders have the same surface area and the same volume. b. The cylinders have the same surface area and different volumes. c. The cylinders have different surface areas and the same volume. d. The cylinders have different surface areas and different volumes. 6. The surface area of a rectangular prism is 2,116 mm2. If you double the length, width, and height of the prism, what will be the surface area of the new prism? How much larger is this amount? 7.
    [Show full text]
  • Why Cells Are Small: Surface Area to Volume Ratios
    Name__________________________________ Why Cells are Small: Surface Area to Volume Ratios Today’s lab is about the size of cells. To understand this, we first have to understand surface area to volume ratios. Let’s start with surface area. Get a box of sugar cubes. The surface area of the cube is calculated by finding the area (length x width) of one side. But we want to know the whole surface. On a cube, there are 6 sides, so we multiply that area x 6. So Surface area of a cube = length x width x 6 Volume is another simple calculation – we just multiply length x width x height. Build the following four structures from your sugar cubes, and fill out the table below. The one we’re most interested in, though is the surface area to volume ratio, which is just Surface area ÷ volume A B C D 1 UNIT 2 UNITS 3 4 UNITS OR CM UNITS Figure Total number Surface area of Volume of Surface area to Total surface of Cubes figure figure Volume Ratio of individual (=6 x length x (= length x ( = SA / V) cubes* width) width x height) (= 6 * # cubes) A 1 6 1 6 1 B 8 24 8 3 48 C 27 54 27 2 162 64 96 64 1.5 384 D *because surface area of one cube is 6 Surface Area to Volume Lab Questions 1. What happened to the surface area as the size increased? It increased. 2. What happened to the volume as the size increased? It also increased, but not as quickly.
    [Show full text]
  • Surface Area of 3-D Objects: Prisms: a Prism Is a Polyhedron with Two Opposite Faces That Are Identical Polygonal Regions Called the Bases
    Surface Area of 3-d Objects: Prisms: A prism is a polyhedron with two opposite faces that are identical polygonal regions called the bases. The vertices of the bases are joined with segments to form the lateral faces which are parallelograms. If the lateral faces are rectangles, the prism is called a right prism, otherwise, it’s called an oblique prism. oblique right The surface area is the sum of the areas of the two bases and the areas of the lateral faces. Right rectangular prism: Right triangular prism: 5 ft 7” 4 ft 5 ft 3 ft 3” 4” 1 2 SA..243 =( )( ) + 247( )( ) + 237( )( ) SA. .=( 3)( 5) + 2 ( 3)( 4) +( 4)( 5) + ( 5) 2 bottom rectangle square top and bottom two sides two sides two triangles 2 =24 + 56 + 42 = 122in =15 + 12 + 20 + 25 = 72 ft 2 A cube has a total surface area of 384 square units. What is the length of one of its edges? The surface area of a cube is 6s2 , so 6s22= 384 s = 64 s = 8 units . If the length of the side of a cube is multiplied by 4, how will the surface area of the larger cube compare to the surface area of the original cube? 2 2 2 6s is the original surface area. 6( 4ss) = 16 6 is the new surface area, so the larger cube has 16 times the original surface area. Pyramids: A pyramid is a polyhedron formed by connecting the vertices of a polygonal region called the base to another point not in the plane of the base called the apex using segments.
    [Show full text]
  • Surface Areas and Volumes
    CHAPTER 13 SURFACE AREAS AND VOLUMES (A) Main Concepts and Results • Cuboid whose length = l, breadth = b and height = h (a) Volume of cuboid = lbh (b) Total surface area of cuboid = 2 ( lb + bh + hl ) (c) Lateral surface area of cuboid = 2 h (l + b) (d) Diagonal of cuboid = lbh222++ • Cube whose edge = a (a) Volume of cube = a3 (b) Lateral Surface area = 4a2 (c) Total surface area of cube = 6a2 (d) Diagonal of cube = a 3 • Cylinder whose radius = r, height = h (a) Volume of cylinder = πr2h (b) Curved surface area of cylinder = 2πrh (c) Total surface area of cylinder = 2πr (r + h) • Cone having height = h, radius = r and slant height = l 1 (a) Volume of cone = πrh2 3 (b) Curved surface area of cone = πrl 16/04/18 122 EXEMPLAR PROBLEMS (c) Total surface area of cone = πr (l + r) (d) Slant height of cone (l) = hr22+ • Sphere whose radius = r 4 (a) Volume of sphere = πr3 3 (b) Surface area of sphere = 4πr2 • Hemisphere whose radius = r 2 (a) Volume of hemisphere = πr3 3 (b) Curved surface area of hemisphere = 2πr2 (c) Total surface area of hemisphere = 3πr2 (B) Multiple Choice Questions Write the correct answer Sample Question 1 : In a cylinder, if radius is halved and height is doubled, the volume will be (A) same (B) doubled (C) halved (D) four times Solution: Answer (C) EXERCISE 13.1 Write the correct answer in each of the following : 1. The radius of a sphere is 2r, then its volume will be 4 8πr3 32 (A) πr3 (B) 4πr3 (C) (D) πr3 3 3 3 2.
    [Show full text]
  • Surface Area of a Sphere in This Example We Will Complete the Calculation of the Area of a Surface of Rotation
    Surface Area of a Sphere In this example we will complete the calculation of the area of a surface of rotation. If we’re going to go to the effort to complete the integral, the answer should be a nice one; one we can remember. It turns out that calculating the surface area of a sphere gives us just such an answer. We’ll think of our sphere as a surface of revolution formed by revolving a half circle of radius a about the x-axis. We’ll be integrating with respect to x, and we’ll let the bounds on our integral be x1 and x2 with −a ≤ x1 ≤ x2 ≤ a as sketched in Figure 1. x1 x2 Figure 1: Part of the surface of a sphere. Remember that in an earlier example we computed the length of an infinites­ imal segment of a circular arc of radius 1: r 1 ds = dx 1 − x2 In this example we let the radius equal a so that we can see how the surface area depends on the radius. Hence: p y = a2 − x2 −x y0 = p a2 − x2 r x2 ds = 1 + dx a2 − x2 r a2 − x2 + x2 = dx a2 − x2 r a2 = dx: 2 2 a − x The formula for the surface area indicated in Figure 1 is: Z x2 Area = 2πy ds x1 1 ds y z }| { x r Z 2 z }| { a2 p 2 2 = 2π a − x dx 2 2 x1 a − x Z x2 a p 2 2 = 2π a − x p dx 2 2 x1 a − x Z x2 = 2πa dx x1 = 2πa(x2 − x1): Special Cases When possible, we should test our results by plugging in values to see if our answer is reasonable.
    [Show full text]