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Module MA1132 (Frolov), Advanced Calculus Homework Sheet 9 Due: at the beginning of the tutorial session Thursday/Friday, 31/32 March 2016 You may use Mathematica to sketch the integration regions and solids, and to check the results of integration. Moment of inertia: The tendency of a solid to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If the solid occupies a region G in an xyz-coordinate system, and if its density function δ(x, y, z) is continuous on G, then the moments of inertia about the x-axis, the y-axis, and the z-axis are denoted by Ix , Iy , and Iz , respectively, and are defined by ZZZ (y 2 + z 2 )δ(x, y, z)dV , Ix = Z Z ZG (x2 + z 2 )δ(x, y, z)dV , Iy = (1) G ZZZ Iz = (x2 + y 2 )δ(x, y, z)dV . G Newton’s law of gravitation: Let a solid occupy a region G in an xyz-coordinate system, and let its density function δ(x, y, z) be continuous on G. Then the gravitational force F = Fx i + Fy j + Fz k exerted by the solid on a point particle of mass m located at (ξ, η, ζ) is given by ZZZ x−ξ δ(x, y, z)dV , Fx (ξ, η, ζ) = Gm r3 G ZZZ y−η Fy (ξ, η, ζ) = Gm δ(x, y, z)dV , r3 (2) G ZZZ z−ζ Fz (ξ, η, ζ) = Gm δ(x, y, z)dV , 3 G r p r = (x − ξ)2 + (y − η)2 + (z − ζ)2 , where G is the gravitational constant. The force can be obtained from the gravitational potential field U (ξ, η, ζ) as follows ZZZ 1 U (ξ, η, ζ) = −G δ(x, y, z)dV , G r ∂U (ξ, η, ζ) ∂U (ξ, η, ζ) ∂U (ξ, η, ζ) Fx (ξ, η, ζ) = −m , Fy (ξ, η, ζ) = −m , Fz (ξ, η, ζ) = −m . ∂ξ ∂η ∂ζ (3) In what follows we set m = 1, G = 1, and consider homogeneous solids with δ(x, y, z) = 1. 1. Considerpthe solid G bounded above by the surface z = h and below by the surface az 2 = h2 x2 + y 2 . 1 (a) What is the surface z = h? (b) Describe the surface az 2 = h2 p x2 + y 2 (c) Sketch the solid G. (d) Sketch the projection of the solid G onto the xy-plane. (e) Find the volume V of the solid G. (f) Find the centroid of the solid G. (g) Find the moments of inertia of the solid G. (h) Find the gravitational force exerted on a point particle by the solid G if the point particle is located at the origin (0, 0, 0). Find its limit as a → ∞ with h kept fixed, and its limit as h → ∞ with a kept fixed. Show the details of your work. Solution: (a) It is a horizontal plane through (0, 0, h) (b) It is a surface of revolution obtained by rotating the parabola az 2 = h2 x in the xz-plane about the √ z-axes. In polar coordinates which we’ll be using it is given by the equation z = √ha r. (c),(d) The solid and its projection R onto the xy-plane are shown below. R is a circle of radius a. (e) The volume V of G is ZZZ ZZ Z V = dV = G R h h √ √ r a Z 2π Z dzdA = 0 0 a h √ ha2 2ha2 1 (h − √ r)rdrdθ = 2π( − ) = πa2 h . 2 5 5 a (4) (f) The x and y coordinates of the centroid are 0 by the symmetry, and the z coordinate is ZZZ Z 2π Z a 1 1 h2 π h2 a2 h2 a2 zc = z dV = (h2 − r)rdrdθ = ( − ) V 2V 0 a V 2 3 G 0 (5) πa2 h2 5 = = h. 6V 6 2 (g) We first find Iz ZZZ 2 Iz = 2π Z 2 Z a Z (x + y ) dV = G 0 2 h √ √ r a 0 a Z r dzrdrdθ = 2π 0 h √ r2 (h − √ r)rdr a (6) 4 4 = 2π( h 2ha π 5 5 ha − ) = ha4 = V a2 = M a2 , 4 9 18 18 18 where V is equal to the mass of the cone because we set δ(x, y, z) = 1. The Ix = Iy = 1 (I + Iy ) by the rotational symmetry, so 2 x Z 2π Z a Z h 1 1 2 2 2 Ix = ( (x + y ) + z ) dV = Iz + z 2 dzrdrdθ √ h 2 √ 0 0 G 2 r a Z a 1 1 3 h3 1 2π a2 h3 2a2 h3 1 π = Iz + 2π (h − 3/2 r3/2 )rdr = Iz + ( − ) = Iz + a2 h3 (7) 2 a 2 3 2 7 2 7 0 3 2 2 h a = 5M ( + ) . 36 7 ZZZ (h) Obviously the only nonvanishing component is Fz Z 2π Z a Z h z z Fz (0, 0, 0) = dV = dzrdrdθ 2 2 2 3/2 h √ (r 2 + z 2 )3/2 √ G (x + y + z ) 0 0 r a Z a Z a Z as 1 r 1 r rdr = 2π q √ −√ dr − 2π = 2π dr . 2 h 2 + h2 2 + h2 h2 r + r r 2 0 0 0 a r + r ZZZ a (8) The first integral is computed by first making the substitution s a r √ = t , r = 0 → t = 0 , r = a → t = 0 a 2 r + ha a2 + h2 h2 t2 , r= a 1 − t2 (9) h2 2t h2 1 dr = dt = , d 2 2 a (1 − t ) a 1 − t2 so that Z 2π 0 a s r h2 dr = 2π 2 a r + ha Z 0 ta 2t h2 t dt = 2π (1 − t2 )2 a Z ta td 0 1 . 1 − t2 Then one uses the integration by parts and gets Z Z h2 ta 1 h2 1 ta h2 ta 1 td = 2π t dt 2π − 2π a 0 1 − t2 a 1 − t2 0 a 0 1 − t2 Z √ h2 ta 1 2 2 = 2π a + h − 2π dt . a 0 1 − t2 3 (10) (11) Finally one uses the partial fraction decomposition and gets Z Z 1 1 h2 1 + t ta h2 ta 1 h2 ta + dt = π ln 2π dt = π a 0 1 − t2 a 0 1+t 1−t a 1−t 0 √ √ h2 a2 + h2 + a h2 a2 + h2 + a =π ln √ = 2π ln a a h a2 + h2 − a Thus, the first integral is equal to √ Z as √ r a2 + h2 + a h2 2 2 ln . 2π a + h − 2π dr = 2π 2 a h r + ha 0 The second integral in (8) gives Z a √ r √ − 2π dr = −2π a2 + h2 − h . r2 + h2 0 (12) (13) (14) Summing up the two contribution one finds h2 Fz (0, 0, 0) = 2πh − 2π ln a √ a2 + h2 + a . h (15) The limits are easily found by using the Taylor expansion r h2 a2 a h2 a2 a Fz (0, 0, 0) = 2πh − 2π ln( 1 + 2 + ) ≈ 2πh − 2π ln(1 + 2 + ) a h h a 2h h (16) 2 2 2 3 3 2 h a a a a a a ≈ 2πh − 2π ( + 2 − 2 − 3 + 3 ) = π → 0 as h → ∞ . a h 2h 2h 2h 3h 3h Interestingly, the force vanishes. Similarly, Fz (0, 0, 0) → 2πh as a → ∞. The force remains finite even though the solid becomes infinitely heavy. 2. Consider the solid G bounded by the surfaces z = 0, z = h and x2 + y 2 = a2 . (a) What is the solid? (b) Find the gravitational force exerted on a point particle by the solid G if the point particle is located on the z-axis at (0, 0, ζ) (ζ can be any real number) (c) Plot its graph and the graph of ∂Fz /∂ζ for a = 1, h = 2 (use Mathematica). (d) Find the limit of Fz as a → ∞ with h and ζ kept fixed. (e) Find the limit of Fz as h → ∞ with a and ζ kept fixed. Solution: (a) It is a circular solid cylinder of radius a and height h with the centre located at the origin and the axis of symmetry coinciding with the z-axis. 4 (b) Obviously the only nonvanishing component is Fz ZZZ Z 2π Z a Z h z−ζ z−ζ Fz (0, 0, ζ) = dV = dzrdrdθ 2 2 2 3/2 2 2 3/2 G (x + y + (z − ζ) ) 0 0 0 (r + (z − ζ) ) ! Z a 1 1 p −p = 2π rdr r2 + (h − ζ)2 r2 + ζ 2 0 p p a2 + ζ 2 − |ζ| − a2 + (h − ζ)2 + |h − ζ| . = 2π (17) (c) The plots of Fz and ∂Fz /∂ζ are shown below (a = 1, h = 2) (d) The limit a → ∞ is easily found by using the Taylor expansion 2πh if ζ≤0 2πh − 4πζ if 0 ≤ ζ ≤ h Fz (0, 0, ζ) → 2π (−|ζ| + |h − ζ|) = −2πh if ζ≥h (18) Thus in this limit the gravitational force is constant outside the solid, and changes linearly with the distance to the boundary when the particle is inside it. (e) The limit h → ∞ gives p 2 2 a + ζ − |ζ| > 0 . Fz (0, 0, ζ) → 2π (19) It is interesting that it is symmetric under ζ → −ζ. Can you give a simple explanation to this result? p 3. Consider the solid G bounded below by the surface z = rα , α > 0, r = x2 + y 2 and above by the plane z = 1. Note that the surface z = r is a cone, and z = r2 is a paraboloid. (a) Sketch the surface z = rα for α = 1/2, α = 1, α = 2, and the projection of the solid G onto the xy-plane. (b) Find the volume V of the solid G, and its limit as α → ∞. (c) Find the centroid of the solid G, and its limit as α → ∞. (d) Find the moments of inertia of the solid G, and its limit as α → ∞. 5 (e) Use Mathematica to find the gravitational force exerted on a point particle by the solid G if the point particle is located at the origin (0, 0, 0). (f) Use Mathematica to plot Fz as a function of α, and find its limit as α → ∞. (g) Explain the limiting values obtained in (b), (c), (d), (f). Show the details of your work. Solution: (a) The surface z = rα and the projection R of the solid G onto the xy-plane are shown below. R is a circle of radius 1. (b) The volume V of G is ZZ Z ZZZ dV = V = R G 1 Z 2π Z Z α (1 − r )rdrdθ = 2π dzdA = rα 1 0 1 (r + rα+1 )dr 0 0 1 1 α = 2π( − )=π , 2 α+2 α+2 lim V = π . (20) α→∞ (c) The x and y coordinates of the centroid are 0 by the symmetry, and the z coordinate is ZZZ Z Z π 1 1 π 1 2α zc = z dV = (1 − r )rdr = (r − r2α+1 )dr V V V 0 0 G π 1 1 π α 2+α (21) = ( − )= = , V 2 2 + 2α 2V 1 + α 2 + 2α 1 lim zc = . α→∞ 2 (d) We first find Iz ZZZ Z 1 Z 1 2 2 α 3 Iz = (x + y ) dV = 2π (1 − r )r dr = 2π (r3 − rα+3 )dr G 0 1 1 α = 2π( − )=π , 4 4+α 8 + 2α π lim Iz = . α→∞ 2 6 0 (22) The Ix = Iy = 21 (Ix + Iy ) by the rotational symmetry, so Z 1Z 1 1 2 1 2 2 z 2 dzrdr ( (x + y ) + z ) dV = Iz + 2π Ix = 2 2 α G Z 1 Z0 1 r 1 1 1 2π = Iz + 2π (1 − r3α )rdr = Iz + (r − r3α+1 )dr 2 2 3 0 0 3 1 2π 1 1 α α α(18 + 7α) = Iz + ( − )=π +π =π , 2 3 2 2 + 3α 4(4 + α) 2 + 3α 4(4 + α)(2 + 3α) 7π . lim Ix = α→∞ 12 ZZZ (23) (e) Obviously the only nonvanishing component is Fz ZZZ Z 1Z 1 z z Fz (0, 0, 0) = dV = 2π dzrdr 2 2 2 3/2 2 2 3/2 G (x + y + z ) 0 rα (r + z ) Z 1 Z 1 1 1 1 r √ √ = 2π −√ rdr = 2π −√ dr r2 + r2α r2 + 1 r2α−2 + 1 r2 + 1 0 0 α−2 √ ; 4−3α ;−1) 2π − 2 F1 ( 21 , 2(α−1) 2−2α − 2+1 if 0 ≤ α ≤ 1 α−2 = , √ 1 1 1 2π 2 F1 2 , 2α−2 ; 1 + 2α−2 ; −1 − 2 + 1 if α≥1 (24) where 2 F1 is a hypergeometric function. (f) The plot of Fz is shown below The limit is √ Fz (0, 0, 0) → 2π 2 − 2 ≈ 3.6806 as α → ∞ . (25) (g) In the limit α → ∞ the solid becomes a cylinder of radius 1 and of height 1. The result in (f) agrees with (17) where one sets a = h = 1 and ζ = 0. 7