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Unformatted text preview: 18.02 HOMEWORK #7, DUE NOVEMBER 18, 2010 BJORN POONEN 1. Part A (After Nov. 4) p. 986: 40* (10 pts.; “the solid” means only the part in the first octant; assume constant density 1) Solution. Let T be the solid. For a fixed z ≥ 0, the range for x and y is the interval [0 , √ 1 z 2 ]. Thus the moment of inertia of T is I := ZZZ T (distance to zaxis) 2 dm = ZZZ T ( x 2 + y 2 ) dV = Z 1 z =0 Z √ 1 z 2 y =0 Z √ 1 z 2 x =0 ( x 2 + y 2 ) dxdy dz = 2 Z 1 z =0 Z √ 1 z 2 y =0 Z √ 1 z 2 x =0 x 2 dxdy dz (by symmetry) = 2 Z 1 z =0 Z √ 1 z 2 y =0 (1 z 2 ) 3 / 2 3 dy dz = 2 3 Z 1 z =0 (1 z 2 ) 2 dz = 2 3 Z 1 z =0 (1 2 z 2 + z 4 ) dz = 2 3 1 2 3 + 1 5 = 16 45 . (After Nov. 4) 5A2b* (10 pts.) Solution. The face of the tetrahedron passing through (1 , , 0), (0 , 2 , 0), and (0 , , 2) is x 1 + y 2 + z 2 = 1 , which can be rewritten as z = 2 2 x y. 1 If we fix x and y , then the range for z is the interval [0 , 2 2 x y ]. This gives the limits for the inner integral, Z 2 2 x y z =0 . The outer integrals are over the triangle bounded by the axes and 2 2 x y = 0, i.e., the triangle with vertices (0 , 0), (1 , 0) and (0 , 2). If we fix x , the range for y is [0 , 2 2 x ]. Finally the range for x is [0 , 1]. So the answer is Z 1 x =0 Z 2 2 x y =0 Z 2 2 x y z =0 ··· dz dy dx. (After Nov. 5) p. 843: 5, 11, 21, 49, 58* (10 pts.) Solution. (to p. 843, 5) We are given r = 2, θ = 1 3 π , z = 5, and we compute x = r cos θ = 2 1 2 = 1 y = r sin θ = 2 √ 3 2 = √ 3 z = z = 5 . So the answer is (1 , √ 3 , 5). Solution. (to p. 843, 11) We are given ρ = 2, φ = 1 3 π , θ = 3 2 π , and we compute x = ρ sin φ cos θ = 2 · √ 3 2 · 0 = 0 y = ρ sin φ sin θ = 2 · √ 3 2 · ( 1) = √ 3 z = ρ cos φ = 2 · 1 2 = 1 . So the answer is (0 , √ 3 , 1). Solution. (to p. 843, 21) We are given x = 3, y = 4, z = 12. The cylindrical coordinates are r = p x 2 + y 2 = √ 3 2 + 4 2 = 5 θ = tan 1 y x = tan 1 4 3 (since (3 , 4) is in the first quadrant) z = z = 12 , 2 i.e., ( r,θ,z ) = (5 , tan 1 4 3 , 12). The spherical coordinates are ρ = p x 2 + y 2 + 2 = √ 3 2 + 4 2 + 12 2 = 13 φ = cos 1 z ρ = cos 1 12 13 θ = tan 1 y x = tan 1 4 3 , since (3 , 4) is in the first quadrant; i.e., ( ρ,φ,θ ) = (13 , cos 1 12 13 , tan 1 4 3 ). Solution. (to p. 843, 49) Here r and z are cylindrical coordinates. In the plane, 1 ≤ r ≤ 3 describes the annulus obtained by starting with the circle of radius 3 centered at the origin, and cutting out the circle of radius 1 centered at the origin. In space, 1 ≤ r ≤ 3 describes an infinite cylinder of radius 3 with an infinite cylinder of radius 1 removed from its center.an infinite cylinder of radius 3 with an infinite cylinder of radius 1 removed from its center....
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This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.
 Fall '08
 Auroux
 Multivariable Calculus

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