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Module MA1132 (Frolov), Advanced Calculus Tutorial Sheet 9 To be solved during the tutorial session Thursday/Friday, 24/25 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 Fx (ξ, η, ζ) = Gm δ(x, y, z)dV , r3 Z Z ZG y−η δ(x, y, z)dV , Fy (ξ, η, ζ) = Gm r3 (2) Z Z ZG 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. Consider the solid G bounded by the surfaces z = 0, z = h and and x2 + y 2 = a2 . (a) What are the surfaces z = 0 and z = h? 1 (b) What is the surface x2 + y 2 = a2 ? (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) They are horizontal planes through (0, 0, 0) and (0, 0, h). (b) It is a circular cylinder (c),(d) The solid and its projection R onto the xy-plane are shown below 1.0 0.5 0.0 -0.5 -1.0 -1.0 0.0 -0.5 0.5 1.0 (e) The volume V of G is ZZZ ZZ Z V = dV = G h Z 2π Z dzdA = R 0 a hrdrdθ = πa2 h . (4) 0 0 (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 π h2 a2 1 (5) zc = z dV = h2 rdrdθ = = h. V 2V 0 V 2 2 G 0 (g) We first find Iz ZZZ 2 Iz = 2 2π Z Z a Z (x + y ) dV = G h 2 Z r dzrdrdθ = 2π 0 ha4 1 1 = 2π = V a2 = M a2 , 4 2 2 2 0 0 0 a r3 hdr (6) 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 z 2 dzrdrdθ Ix = ( (x + y ) + z ) dV = Iz + 2 2 0 0 0 G Z a 1 3 1 1 2π a2 h3 a2 h2 = Iz + 2π h rdr = Iz + = M( + ) . 2 2 3 2 4 3 0 3 ZZZ (h) Obviously the only nonvanishing component is Fz ZZZ Z 2π Z a Z h z z Fz (0, 0, 0) = dV = dzrdrdθ 2 2 2 3/2 2 2 3/2 0 0 (r + z ) G (x + y + z ) 0 Z a Z a r 1 1 1− √ −√ rdr = 2π dr = 2π r r2 + h2 r2 + h2 0 0 √ √ = 2πa − 2π a2 + h2 − h = 2π(a + h − a2 + h2 ) . (7) (8) It is interesting that the result is symmetric under a ↔ h. The limits are easily found by using the Taylor expansion r h2 h2 Fz (0, 0, 0) = 2π(a + h − a 1 + 2 ) ≈ 2π(a + h − a(1 + 2 )) → 2πh as a → ∞ . (9) a 2a Similarly, Fz (0, 0, 0) → 2πa as h → ∞. The force remains finite even though the solid becomes infinitely heavy. 2. Consider the solid G bounded above by the surface 15 p 2 1 − x + y 2 and below by the surface z = (x2 + y 2 ) . 2 2 p (a) What is the surface z = 15 − x2 + y 2 ? 2 z= (b) What is the surface z = 21 (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 mass M of the solid G if its density is √ 2 2 e x +y − 1 p δ(x, y, z) = . x2 + y 2 + 5 x2 + y 2 (g) What is the density of the solid G at the origin: δ(0, 0, 0) =? Show the details of your work. Solution: (a) It is a cone 3 (b) It is a paraboloid (c),(d) The solid and its projection R onto the xy-plane are shown below 3 2 1 0 -1 -2 -3 -3 -2 -1 0 1 2 R is a circle of radius 3 as one can find by solving the equation 3 15 2 − r − 12 r2 = 0. (e) The volume V of G is Z 2π Z 3 ZZZ 15 1 135 81 117 ( − r − r2 )rdrdθ = 2π( dV = −9− )= π. V = 2 2 4 8 4 0 0 G (f) The mass M of G is Z 2π Z 3 r ZZZ e − 1 15 1 δ(x, y, z)dV = ( − r − r2 )rdrdθ M= 2 r + 5r 2 2 0 0 Z 3G 17 (er − 1)(3 − r)dr = e3 − =π π ≈ 36.397 . 2 0 (g) δ(0, 0, 0) = 1/5 4 (10) (11)