DOCSLIB.ORG

Calculus 2 Tutor Worksheet 12 Surface Area of Revolution In

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Calculus 2 Tutor Worksheet 12 Surface Area of Revolution In Calculus 2 Tutor Worksheet 12 Surface Area of Revolution in Parametric Equations Worksheet for Calculus 2 Tutor, Section 12: Surface Area of Revolution in Parametric Equations 1. For the function given by the parametric equation x = t; y = 1: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape. c 2018 MathTutorDVD.com 1 2. For the function given by the parametric equation x = t; y = 2t + 1: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is the surface of rotation in geometric terms? Compare the results of the above questions to the geometric formula for the surface area of this shape. 3. For the function given by the parametric equation x = 2t + 1; y = t: (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = 1; c 2018 MathTutorDVD.com 2 (b) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = T for any T > 0; (c) What is the Cartesian equation for this function? (d) What is this function in geometric terms? Compare the results of the above ques- tions to the geometric formula for the surface area of this shape. 4. For the function given by the parametric equation x = t; y = −at + 1 for a > 0: 1 (a) Find the surface area of f rotated about the x-axis as t goes from t = 0 to t = a ; c 2018 MathTutorDVD.com 3 1 (b) What is the above surface area (from t = 0 to t = a ) as a ! 1? Explain your results in geometric terms. (c) Challenge. Parametrize the same function as x = cos2 θ; y = −a cos2 θ + 1. Find π the surface area of f rotated about the x-axis as θ goes from 0 to 2 . Assume a ≤ 1. Explain your results geometrically, especially considering the case a = 1. Hint: to simplify, use the expressions cos 2θ = 2 cos2 θ − 1; sin 2θ = 2 sin θ cos θ: 5. Find the surface area of a sphere by rotating the unit circle around the x-axis: p (a) The unit circle (the upper semi-circle) can be parametrized as x = t; y = 1 − t2; −1 ≤ t ≤ 1. Rotate this parametrization about the x-axis to find the surface area of the unit sphere. c 2018 MathTutorDVD.com 4 (b) The upper semi-circle of the unit circle can also be parametrized as x = cos θ; y = sin θ; 0 ≤ θ ≤ π. Rotate this parametrization about the x-axis. to find the surface area of the unit sphere. (c) Find the surface area of the sphere of radius r by rotating x = r cos θ; y = r sin θ about the x-axis. 6. Find the surface area of the following surfaces of revolution: p (a) The function x = t; y = 2 t between t = 0 and 9. p p 2 (b) The function x = 2 t; y = 3 t t between t = 0 and t = 4. c 2018 MathTutorDVD.com 5 p p 2 (c) The function x = 3 t t; y = 2 t between t = 0 and t = 1. t3 2 (d) The function x = 3 − t; y = t between t = 2 and t = 5. p 2 t3 (e) The function x = 2t ; y = 3 − 2t between t = 3 and t = 4. 7. Find the surface area of the following surfaces of revolution: p (a) The function x = sin−1 t; y = 1 − t2 between t = −1 and t = 1. What is this function in Cartesian terms? c 2018 MathTutorDVD.com 6 p (b) The function x = sinh−1 t; y = 1 + t2 between t = −1 and t = 1. What is this function in Cartesian terms? (c) The function x = ln t; y = t between t = 1 and 2. (d) The function x = sech t; y = tanh t from t = 0 to t = 1. 8. Find the surface area of the following surfaces of revolution: 2 π (a) The function x = cos θ; y = sin θ between θ = 0 and θ = 2 . 1 2 π (b) The function x = 2 sec θ; y = sec θ between θ = 0 and θ = 4 . Hint: make a c 2018 MathTutorDVD.com 7 u-substitution related to a derivative that you computed earlier in this problem. c 2018 MathTutorDVD.com 8 Answer key. 1. The parametric equation x = t; y = 1: 1(a). Answer: 2π . To find the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Here since x(t) = t, dx = 1 dt Since y(t) = 1, dy = 0 dt Then, we compute Z t=1 p Z 1 A = 2π · 1 · 12 + 02dt = 2πdt t=0 0 which is 2π . c 2018 MathTutorDVD.com 9 1(b). Answer: 2πT . To find the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Here, x0(t) = 1; y0(t) = 0 as in the last problem. However, the bounds α and β are different. Before α = 0; β = 1, but now α = 0; β = T . Then the surface area is Z T p Z T A = 2π1 12 + 02dt = 2πdt 0 0 which is 2πT . c 2018 MathTutorDVD.com 10 1(c). Answer: y = 1 . If x = t and y = 1, this means that x can be anything but y is always 1. This is a horizontal line. Between t = 0 and 1, then x can vary from 0 to 1, which is a line segment of length 1. Between t = 0 and T , then x can vary from 0 to T , which is a line segment of length T . 1(d). Answer: a cylinder . Rotating the line y = 1 (between any two bounds of x) about the x-axis yields a cylinder. The surface area of a cylinder is 2πr2 + 2πrh However, the 2πr2 accounts for the circles on the top and bottom. The surface area of the proper surface of rotation (the “face” of the cylinder) is 2πrh The line y = 1 rotated about the x-axis (that is, y = 0) has radius 1, so r = 1 The height is the distance between x = 0 and x = T , which is h = T Therefore, the surface area is 2πT which matches the result we derived for the surface area of this surface of rota- tion. c 2018 MathTutorDVD.com 11 2. The parametric equation x = t; y = 2t + 1: p 2(a). Answer: 4π 5 . To find the surface area of this function rotated about the x-axis, we compute: s Z β dx2 dy 2 A = 2πy(t) + dt α dt dt Since x(t) = t, then dx = 1 dt Since y(t) = 2t + 1, then dy = 2 dt The bounds of t are between 0 and 1, so α = 0 and β = 1. Then, Z 1 p Z 1 A = 2π (2t + 1) p(12 + 22)dt = 2π 5 2t + 1dt 0 0 According to the technique of Calculus I, this integral is Z 2t + 1dt = t2 + t + C so we have p p 2 1 A = 2π 5 t + t 0 = 2π 5 (2 − 0) p which is 4π 5 . c 2018 MathTutorDVD.com 12 p 2(b). Answer: 2π 5 T 2 + T . This is the same integral as the last problem but with a difference in the boundaries of integration. The integral is Z T p p Z T p 2 2 2 T A = 2π(2t + 1) 1 + 2 dt = 2π 5 2t + 1dt = 2π 5 t + t 0 0 0 p which is 2π 5 T 2 + T . 2(c). Answer: y = 2x + 1 . Substituting x = t into y = 2t + 1 gives the answer y = 2x + 1 . Between t = 0 and t = 1 (that is, between x = 0 and x = t), this is the line between the points (0; 1) ! (1; 3) Between t = 0 and t = T (that is, between x = 0 and x = T ), this is the line between the points (0; 1) ! (T; 2T + 1) c 2018 MathTutorDVD.com 13 2(d). Answer: a cone segment or truncated cone. Rotating the line segment from (0; 1) to (1; 3) about the x-axis yields a truncated cone (cone segment). Rotating 1 the longer line segment from − 2 ; 0 to (1; 3) would be an entire cone, but since the line segment is shorter, the cone is truncated. The surface area of the face of a truncated cone (not counting the base circles) is p 2 2 π(r2 + r1) h + (r2 − r1) where r2 is the larger radius, r1 is the smaller radius, and h is the height. The smaller radius r1 is just r1 = 1 since it is the circle made by rotating the point (0; 1) around the x-axis.
Load more

Recommended publications
  • Parametric Surfaces and 16.6 Their Areas Parametric Surfaces
    Parametric Surfaces and 16.6 Their Areas Parametric Surfaces 2 Parametric Surfaces Similarly to describing a space curve by a vector function r(t) of a single parameter t, a surface can be expressed by a vector function r(u, v) of two parameters u and v. Suppose r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k is a vector-valued function defined on a region D in the uv-plane. So x, y, and, z, the component functions of r, are functions of the two variables u and v with domain D. The set of all points (x, y, z) in R3 s.t. x = x(u, v), y = y(u, v), z = z(u, v) and (u, v) varies throughout D, is called a parametric surface S. Typical surfaces: Cylinders, spheres, quadric surfaces, etc. 3 Example 3 – Important From Book The vector equation of a plane through (x0, y0, z0) and containing vectors is rather 4 Example – Point on Surface? Does the point (2, 3, 3) lie on the given surface? How about (1, 2, 1)? 5 Example – Identify the Surface Identify the surface with the given vector equation. 6 Example – Identify the Surface Identify the surface with the given vector equation. 7 Example – Find a Parametric Equation The part of the hyperboloid –x2 – y2 + z = 1 that lies below the rectangle [–1, 1] X [–3, 3]. 8 Example – Find a Parametric Equation The part of the cylinder x2 + z2 = 1 that lies between the planes y = 1 and y = 3. 9 Example – Find a Parametric Equation Part of the plane z = 5 that lies inside the cylinder x2 + y2 = 16.
    [Show full text]
  • Multivariable and Vector Calculus
    Multivariable and Vector Calculus Lecture Notes for MATH 0200 (Spring 2015) Frederick Tsz-Ho Fong Department of Mathematics Brown University Contents 1 Three-Dimensional Space ....................................5 1.1 Rectangular Coordinates in R3 5 1.2 Dot Product7 1.3 Cross Product9 1.4 Lines and Planes 11 1.5 Parametric Curves 13 2 Partial Differentiations ....................................... 19 2.1 Functions of Several Variables 19 2.2 Partial Derivatives 22 2.3 Chain Rule 26 2.4 Directional Derivatives 30 2.5 Tangent Planes 34 2.6 Local Extrema 36 2.7 Lagrange’s Multiplier 41 2.8 Optimizations 46 3 Multiple Integrations ........................................ 49 3.1 Double Integrals in Rectangular Coordinates 49 3.2 Fubini’s Theorem for General Regions 53 3.3 Double Integrals in Polar Coordinates 57 3.4 Triple Integrals in Rectangular Coordinates 62 3.5 Triple Integrals in Cylindrical Coordinates 67 3.6 Triple Integrals in Spherical Coordinates 70 4 Vector Calculus ............................................ 75 4.1 Vector Fields on R2 and R3 75 4.2 Line Integrals of Vector Fields 83 4.3 Conservative Vector Fields 88 4.4 Green’s Theorem 98 4.5 Parametric Surfaces 105 4.6 Stokes’ Theorem 120 4.7 Divergence Theorem 127 5 Topics in Physics and Engineering .......................... 133 5.1 Coulomb’s Law 133 5.2 Introduction to Maxwell’s Equations 137 5.3 Heat Diffusion 141 5.4 Dirac Delta Functions 144 1 — Three-Dimensional Space 1.1 Rectangular Coordinates in R3 Throughout the course, we will use an ordered triple (x, y, z) to represent a point in the three dimensional space.
    [Show full text]
  • Math 1300 Section 4.8: Parametric Equations A
    MATH 1300 SECTION 4.8: PARAMETRIC EQUATIONS A parametric equation is a collection of equations x = x(t) y = y(t) that gives the variables x and y as functions of a parameter t. Any real number t then corresponds to a point in the xy-plane given by the coordi- nates (x(t); y(t)). If we think about the parameter t as time, then we can interpret t = 0 as the point where we start, and as t increases, the parametric equation traces out a curve in the plane, which we will call a parametric curve. (x(t); y(t)) • The result is that we can \draw" curves just like an Etch-a-sketch! animated parametric circle from wolfram Date: 11/06. 1 4.8 2 Let's consider an example. Suppose we have the following parametric equation: x = cos(t) y = sin(t) We know from the pythagorean theorem that this parametric equation satisfies the relation x2 + y2 = 1 , so we see that as t varies over the real numbers we will trace out the unit circle! Lets look at the curve that is drawn for 0 ≤ t ≤ π. Just picking a few values we can observe that this parametric equation parametrizes the upper semi-circle in a counter clockwise direction. (0; 1) • t x(t) y(t) 0 1 0 p p π 2 2 •(1; 0) 4 2 2 π 2 0 1 π -1 0 Looking at the curve traced out over any interval of time longer that 2π will indeed trace out the entire circle.
    [Show full text]
  • Parametric Curves You Should Know
    Parametric Curves You Should Know Straight Lines Let a and c be constants which are not both zero. Then the parametric equations determining the straight line passing through (b; d) with slope c=a (i.e., the line y − d = c=a(x − b)) are: x(t) = at + b ; −∞ < t < 1: y(t) = ct + d Note that when c = 0, the line is horizontal of the form y = d, and when a = 0, the line is vertical of the form x = b. 15 10 x(t)= 2t+ 3, y(t)=t+4 5 x(t)=3-2t, y(t)= 2t+1 �������� x(t)=t+ 2, y(t)=3- 3t x(t)= 3, y(t)=2+ 3t -10 -5 5 10 15 x(t)=2+3t, y(t)=3 -5 -10 Figure 1 A collection of lines whose parametric equations are given as above. Circles Let r > 0 and let x0 and y0 be real numbers. Then the parametric equations determining the circle of radius r centered at (x0; y0) are: x(t) = r cos t + x 0 ; 0 < t < 2π: (1) y(t) = r sin t + y0 To get a circular arc instead of the full circle, restrict the t-values in (1) to t1 < t < t2. 1 1 1 2 3 4 5 2 3 4 5 (3,-1) -1 -1 (3,-1) -2 -2 -3 -3 (a) 0 ≤ t ≤ 2π (b) π=4 ≤ t ≤ 7π=6 Figure 2 The circle x(t) = 2 cos t + 3, y(t) = 2 sin t − 1 and a circular arc thereof. Note that the center is at (3; 1).
    [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]
  • Polynomial Curves and Surfaces
    Polynomial Curves and Surfaces Chandrajit Bajaj and Andrew Gillette September 8, 2010 Contents 1 What is an Algebraic Curve or Surface? 2 1.1 Algebraic Curves . .3 1.2 Algebraic Surfaces . .3 2 Singularities and Extreme Points 4 2.1 Singularities and Genus . .4 2.2 Parameterizing with a Pencil of Lines . .6 2.3 Parameterizing with a Pencil of Curves . .7 2.4 Algebraic Space Curves . .8 2.5 Faithful Parameterizations . .9 3 Triangulation and Display 10 4 Polynomial and Power Basis 10 5 Power Series and Puiseux Expansions 11 5.1 Weierstrass Factorization . 11 5.2 Hensel Lifting . 11 6 Derivatives, Tangents, Curvatures 12 6.1 Curvature Computations . 12 6.1.1 Curvature Formulas . 12 6.1.2 Derivation . 13 7 Converting Between Implicit and Parametric Forms 20 7.1 Parameterization of Curves . 21 7.1.1 Parameterizing with lines . 24 7.1.2 Parameterizing with Higher Degree Curves . 26 7.1.3 Parameterization of conic, cubic plane curves . 30 7.2 Parameterization of Algebraic Space Curves . 30 7.3 Automatic Parametrization of Degree 2 Curves and Surfaces . 33 7.3.1 Conics . 34 7.3.2 Rational Fields . 36 7.4 Automatic Parametrization of Degree 3 Curves and Surfaces . 37 7.4.1 Cubics . 38 7.4.2 Cubicoids . 40 7.5 Parameterizations of Real Cubic Surfaces . 42 7.5.1 Real and Rational Points on Cubic Surfaces . 44 7.5.2 Algebraic Reduction . 45 1 7.5.3 Parameterizations without Real Skew Lines . 49 7.5.4 Classification and Straight Lines from Parametric Equations . 52 7.5.5 Parameterization of general algebraic plane curves by A-splines .
    [Show full text]
  • Lecture 1: Explicit, Implicit and Parametric Equations
    4.0 3.5 f(x+h) 3.0 Q 2.5 2.0 Lecture 1: 1.5 Explicit, Implicit and 1.0 Parametric Equations 0.5 f(x) P -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.5 1.0 1.5 2.0 2.5 3.0 3.5 x x+h -0.5 -1.0 Chapter 1 – Functions and Equations Chapter 1: Functions and Equations LECTURE TOPIC 0 GEOMETRY EXPRESSIONS™ WARM-UP 1 EXPLICIT, IMPLICIT AND PARAMETRIC EQUATIONS 2 A SHORT ATLAS OF CURVES 3 SYSTEMS OF EQUATIONS 4 INVERTIBILITY, UNIQUENESS AND CLOSURE Lecture 1 – Explicit, Implicit and Parametric Equations 2 Learning Calculus with Geometry Expressions™ Calculus Inspiration Louis Eric Wasserman used a novel approach for attacking the Clay Math Prize of P vs. NP. He examined problem complexity. Louis calculated the least number of gates needed to compute explicit functions using only AND and OR, the basic atoms of computation. He produced a characterization of P, a class of problems that can be solved in by computer in polynomial time. He also likes ultimate Frisbee™. 3 Chapter 1 – Functions and Equations EXPLICIT FUNCTIONS Mathematicians like Louis use the term “explicit function” to express the idea that we have one dependent variable on the left-hand side of an equation, and all the independent variables and constants on the right-hand side of the equation. For example, the equation of a line is: Where m is the slope and b is the y-intercept. Explicit functions GENERATE y values from x values.
    [Show full text]
  • Plotting Circles and Ellipses Parametrically Example 1: the Unit Circle
    Lesson 1 Plotting Circles and Ellipses Parametrically Example 1: The Unit Circle Let's compare traditional and parametric equations for the unit circle : Traditional : Parametric : x22 y 1 x(t) cos(t) y(t) sin(t) P(t) cos(t),sin(t) t is called a parameter 0 t 2 You can see that the parametric equation satisfies the traditional equation by substituting one into the other: x22 y 1 (cos(t))22 (sin(t)) 1 11 Created by Christopher Grattoni. All rights reserved. Example 2: Circle of Radius r Let's compare traditional and parametric equations for a circle of radius r centeredTraditional at the : origin : Parametric : 2 2 2 x y r x(t) r cos(t) y(t) r sin(t) P(t) r cos(t),sin(t) 0 t 2 Orientation: Counterclockwise Think of this as dilating the unit circle by a factor of r. Created by Christopher Grattoni. All rights reserved. Example 3: Recentering the Circle Let's compare traditional and parametric equations for a circle of radius r centeredTraditional at (h,k) : : Parametric : 2 2 2 (x h) (y k) r x(t) r cos(t) h y(t) r sin(t) k P(t) r cos(t),sin(t) (h,k) 0 t 2 Think of this as dilating the unit circle by a factor of r and translating by the point (h,k). Let's add the orientation: Created by Christopher Grattoni. All rights reserved. Example 4: Ellipses Let's compare traditional and parametric equations for an ellipse centeredTraditional at (h,k) : : Parametric : 2 2 xh yk x(t) acos(t) h 1 ab y(t) bsin(t) k P(t) acos(t),bsin(t) (h,k) 0 t 2 Think of this as dilating the unit circle by a factor of "a" in the x-direction, a factor of "b" in the y-direction, and translated by the Created by Christopherpoint Grattoni.
    [Show full text]
  • Parametric Curves in the Plane
    Parametric Curves in the Plane Purpose The purpose of this lab is to introduce you to curve computations using Maple for para- metric curves and vector-valued functions in the plane. Background By parametric curve in the plane, we mean a pair of equations x = f(t) and y = g(t) for t in some interval I. A vector-valued function in the plane is a function r(t) that associates a vector in the plane with each value of t in its domain. Such a vector valued function can always be written in component form as follows, r(t) = f(t)i + g(t)j where f and g are functions defined on some interval I. From our definition of a para- metric curve, it should be clear that you can always associate a parametric curve with a vector-valued function by just considering the curve traced out by the head of the vector. Defining parametric curves and vector valued functions simply in Maple The easiest way to define a vector function or a parametric curve is to use the Maple list notaion with square brackets[]. Strictly speaking, this does not define something that Maple recognizes as a vector, but it will work with all of the commands you need for this lab. >f:=t->[2*cos(t),2*sin(t)]; You can evaluate this function at any value of t in the usual way. >f(0); This is how to access a single component. You would use f(t)[2] to get the second component. >f(t)[1] Plotting and animating curves in the plane The ParamPlot command is in the CalcP package so you have to load it first.
    [Show full text]
  • Radius-Of-Curvature.Pdf
    CHAPTER 5 CURVATURE AND RADIUS OF CURVATURE 5.1 Introduction: Curvature is a numerical measure of bending of the curve. At a particular point on the curve , a tangent can be drawn. Let this line makes an angle Ψ with positive x- axis. Then curvature is defined as the magnitude of rate of change of Ψ with respect to the arc length s. Ψ Curvature at P = It is obvious that smaller circle bends more sharply than larger circle and thus smaller circle has a larger curvature. Radius of curvature is the reciprocal of curvature and it is denoted by ρ. 5.2 Radius of curvature of Cartesian curve: ρ = = (When tangent is parallel to x – axis) ρ = (When tangent is parallel to y – axis) Radius of curvature of parametric curve: ρ = , where and – Example 1 Find the radius of curvature at any pt of the cycloid , – Solution: Page | 1 – and Now ρ = = – = = =2 Example 2 Show that the radius of curvature at any point of the curve ( x = a cos3 , y = a sin3 ) is equal to three times the lenth of the perpendicular from the origin to the tangent. Solution : – – – = – 3a [–2 cos + ] 2 3 = 6 a cos sin – 3a cos = Now = – = Page | 2 = – – = – – = = 3a sin …….(1) The equation of the tangent at any point on the curve is 3 3 y – a sin = – tan (x – a cos ) x sin + y cos – a sin cos = 0 ……..(2) The length of the perpendicular from the origin to the tangent (2) is – p = = a sin cos ……..(3) Hence from (1) & (3), = 3p Example 3 If & ' are the radii of curvature at the extremities of two conjugate diameters of the ellipse = 1 prove that Solution: Parametric equation of the ellipse is x = a cos , y=b sin = – a sin , = b cos = – a cos , = – b sin The radius of curvature at any point of the ellipse is given by = = – – – – – Page | 3 = ……(1) For the radius of curvature at the extremity of other conjugate diameter is obtained by replacing by + in (1).
    [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]
  • Villarceau-Section’ to Surfaces of Revolution with a Generating Conic
    Journal for Geometry and Graphics Volume 6 (2002), No. 2, 121{132. Extension of the `Villarceau-Section' to Surfaces of Revolution with a Generating Conic Anton Hirsch Fachbereich Bauingenieurwesen, FG Stahlbau, Darstellungstechnik I/II UniversitÄat Gesamthochschule Kassel Kurt-Wolters-Str. 3, D-34109 Kassel, Germany email: [email protected] Abstract. When a surface of revolution with a conic as meridian is intersected with a double tangential plane, then the curve of intersection splits into two con- gruent conics. This decomposition is valid whether the surface of revolution inter- sects the axis of rotation or not. It holds even for imaginary surfaces of revolution. We present these curves of intersection in di®erent cases and we also visualize imaginary curves. The arguments are based on geometrical reasoning. But we also give in special cases an analytical treatment. Keywords: Villarceau-section, ring torus, surface of revolution with a generating conic, double tangential plane MSC 2000: 51N05 1. Introduction Due to Y. Villarceau the following statement it is valid (compare e.g. [1], p. 412, [3], p. 204, or [4]): The curve of intersection between a ring torus ª and any double tangential plane ¿ splits into two congruent circles. We assume that r is the radius of the meridian circles k of ª and that their centers are in the distance d, d > r, from the axis a of rotation. We generalize and replace k by a conic which may also intersect the axis a. Under these conditions it is still true that the intersection with a double tangential plane ¿ is reducible.
    [Show full text]