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MIT 18 01 - Lecture Notes

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18.01 Calculus Jason Starr Fall 2005 Lecture 22. November 4, 2005 Homework. Problem Set 6 Part I: (f)–(h); Part II: Problem 2 (a) and (c). Practice Problems. Course Reader: 4G-1, 4G-4, 4G-6, 4H-1, 4H-3. 1. Surface area of a right circular cone. Before attacking the general problem of the surface area of a surface of revolution, consider the simplest case of the area of a right circular cone of base radius R and height H. The slant height of the cone is the length of any line segment from the vertex to a point on the base circle. By the Pythagorean theorem, the slant height S is, S = √R2 + H2 . Imagine the cone is made of paper. Make an incision along a line segment from the vertex to the base circle. The resulting piece of paper may be unfolded to form a sector of a circle. The radius of the sector is the slant height s. The circumference of the sector is the circumference of the original base circle 2πr. The formula for the area and circumference of a sector of a circle give the identity, 1 Area of sector = (Radius of sector) × (Circumference of sector). 2 Thus, the area of the cone equals, 1 .πRS A = (S)(2πR) = 2 In particular, the height H is involved only indirectly (as H depends on H). Next, consider a conical band obtained from a right circular cone of base radius R1 and slant height S1 by removing the the top part of the cone of base radius R2 and slant height S2. In particular, the slant height of the band is the difference, s = S2 − S1,� 18.01 Calculus Jason Starr Fall 2005 and the average radius of the band is the average of R1 and R2, 1 r = (R1 + R2). 2 By similar triangles, S1 S2 = . R1 R2 Rearranging gives, R2S1 = R1S2. Using the formula above, the area of the large cone is, A1 = πR1S1, and the area of the small cone is, A2 = πR2S2. The area A of the band is the difference, A = A1 − A2 = π(R1S1 − R2S2). Since R2S1 equals R1S2, the formula is unchanged by adding πR2S1 and subtracting πR1S2 to get, A = π(R1S1 − R2S2) + π(R2S1 − R1S2) = π((R1 + R2)S1 − (R1 + R2)S2). Simplifying and substituting R1 + R2 = 2r and S1 − S2 = 2 gives, A = 2 .πrs2. Surface area of a surface of revolution. Given a segment of a parametric curve, x = x(t), y = y(t) a ≤ t ≤ b the surface of revolution is the surface obtained by revolving the segment through xyz-space about the y-axis. What is the area of this surface? The answer is called the surface area. The method for computing the surface area is so close to the method for computing the arc length of the curve, the details will be skipped. What is relevant is the differential element of surface area. Given a small interval from t to t + dt, approximate the segment of the parametric curve as a line segment. The surface obtained by revolving a line segment is precisely a band of a cone. The average radius of the cone r is x(t). The slant height of the cone is ds. Thus the area of the band is, dA = 2 = 2 (t) �� dx dt �2 + � dy dt �2 dt.πrds πx� � � � � � � � � 18.01 Calculus Jason Starr Fall 2005 Integrating gives the formula for the surface area of the surface of revolution, A = � dA = �t=b t=a 2 (t) �� dx dt �2 + �dy dt �2 dt.πxExamples. A. Consider the line segment connecting the point (0, H) to the point (R, 0). This has equation, H y = (R − x), 0 ≤ x ≤ R. R The slant height of the line s egment is, S = √R2 + H2 , and the differential arc length of the line segment is, S ds = dx. R Thus the differential element of surface area is, S dA = 2πrds = 2πx dx. R Integrating gives, Rx=R 22πS 2πS x= .πRSA = dA = xdx = R R 2x=0 0 This is the same formula obtained above by more elementary means. B. Consider the parametrized semicircle of radius R in the first and third quadrants, x = R cos(θ), −π π . y = R sin(θ)2 ≤ θ ≤ 2 Revolving about the y-axis gives the sphere of radius R. Thus the surface area of the surface of revolution is the surface area of the sphere of radius R. As computed in the previous lecture, the differential element of arc length is, ds = Rdθ. Thus the differential element of surface area is, dA = 2πrds = 2πx(θ)(Rdθ) = 2π(R cos(θ))(Rdθ) = 2πR2 cos(θ)dθ. Integrating gives, � θ=π/2 π/2A = dA = θ=−π/2 2πR2 cos(θ)dθ = 2πR2 (sin(θ)|−π/2 .� �� � 18.01 Calculus Jason Starr Fall 2005 This evaluates to, A = 2πR2(2) = 42 .πRThe fastest way to remember this is to observe the surface area A and the volume V of a sphere of radius R are related by, dV d A = 4πR2 = = (4πR3/3). dr dr C. An astroid is a curve, 2/3 + y 2/3 2/3 x = a . The part of the astroid in the first quadrant has parametric equation, x = a cos3(t),π . y = a sin3(t)0 ≤ t ≤ 2 The derivatives are, dx = −3a cos 2(t) sin(t), dy = 3a sin2(t) cos(t). dt dt Thus, � �2 � �2dx dy 2+ = (−3a cos (t) sin(t))2 + (3a sin2(t) cos(t))2 = 9a 2 sin2(t) cos 2(t)(cos 2(t) + sin2(t)). dt dt The square root is, � �2 � �2 �dx dy+ = 9a2 sin2(t) cos2(t) = 3a sin(t) cos(t). dt dt So the differential element of arc length is, ds = 3a sin(t) cos(t)dt. Thus the differential element of surface area of the surface of revolution is, 3 4dA = 2πrds = 2πx(t)ds = 2π(a cos (t))(3a sin(t) cos(t))dt = 6πa 2 cos (t) sin(t)dt. Integrating, the surface area is, t=π/2 4A = dA = 6πa 2 cos (t) sin(t)dt. t=0 Subsitute u = cos(t), du = − sin(t)dt, u(0) = 1, u(π/2) = 1 to get, � u=0 � u=1 4A = 6πa 2 u (−du) = 6πa 2 u 4du. u=1 u=0� �� � � � � 18.01 Calculus Jason Starr Fall 2005 Thus the surface area of the surface of revolution is, 1 = 0 62/5.πa5A = 6πa 2 u /5 3. Polar coordinate curves. After the explicit, Cartesian form of a curve as a graph, y = f(x), the next most common representation is us ing polar coordinates. Given a function r = r(θ) and an interval a ≤ θ ≤ b, the associated polar coordinate curve is the parametric curve, x = r(θ) cos(θ), y = r(θ) sin(θ) a ≤ θ ≤ b. For each point on the curve, the distance of the point from the origin is, …


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MIT 18 01 - Lecture Notes

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