ps10sol

# ps10sol - 18.02 Fall 2008 Problem Set 10 Part B Solutions 1...

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Unformatted text preview: 18.02 Fall 2008 Problem Set 10, Part B Solutions 1. div F = x x x2 + y 2 + y y x2 + y 2 = (x2 + y 2 ) - 2x2 (x2 + y 2 ) - 2y 2 + = 0. (x2 + y 2 )2 (x2 + y 2 )2 Ca So the flux out of the circle Ca around (1, 0) of radius a < 1 is zero, because Ra div F dA = 0, where Ra is the disk enclosed by Ca . ^ F n ds = Next, consider a > 1. The previous argument does not work because Ra contains the origin. To avoid the origin, we remove it: consider the region R(a, b) = Ra - Db , the disk Ra with the disk Db of radius b around the origin removed. Let Tb be the circle of radius b around the origin, oriented counterclockwise. For b sufficiently small, Tb is inside of Ra and the oriented boundary of R(a, b) is Ca - Tb (see figure below). By Green's theorem, Ca ^ F n ds - Tb ^ F n ds = div F dA = 0 R(a,b) (One can apply Green's theorem because F and div F are defined in R(a, b).) Moreover, for every b > 0, 1 1 ^ ds = (2b) = 2 F n ds = b b Tb Tb ^ ^ because on Tb , n = x, y /b and F n = x/b2 , y/b2 x, y /b = (x2 + y 2 )/b3 = 1/b. Hence, for a > 1, Ca ^ F n ds = Tb ^ F n ds = 2. Ra Ca (a < 1) Tb R(a, b) Ca (a > 1) 1 0 1 0 0 1-z 2 0 1-z 2 0 1-z 2 1-z 2 1 2. M = 0 dxdydz = 0 (1 - z 2 )dz = 2/3. 3 2 1 0 1 z= M 1 x= M = 3 4 0 zdxdydz = (z - z 3 )dz = 1 0 1-z 2 3 . 8 1 0 1 0 1-z 2 0 1-z 2 3 xdxdydz = 2 /2 0 0 1 - z2 dydz 2 (z = sin , dz = cos d) (1 - z 2 )3/2 dz = 3 4 cos4 d = 9 3 3 = 4 16 64 from Notes 3B or using double angle formulas twice. By symmetry with respect to the plane x = y, x = y . Thus the centroid is 9 9 3 (x, y, z) = . , , 64 64 8 1 3. In cylindrical coordinates, 1 d= 4a3 /3 3 = 4a3 = = 1 4a3 0 2a 0 2a 0 2a 0 2a 0 2 0 2 0 2 0 a2 -(z-a)2 1 2 (r + z 2 )3/2 3 r= r2 + z 2 r drddz 2 2 a -(z-a) ddz r=0 ((2az)3/2 - z 3 ) ddz 1 2a3 1 = 3 2a ((2az)3/2 - z 3 ) dz = 6 a. 5 2 1 (2a)4 - (2a)4 5 4 In spherical coordinates, d = , dV = 2 sin ddd, d= = 3 4a3 3 4a3 2 0 2 0 2 0 0 0 /2 /2 0 2a cos 2 sin ddd 1 (2a cos )4 sin dd 4 /2 3 = 4a3 = 1 -1 (2a)4 (cos )5 4 5 d 0 6a 3 8a4 = . 4a3 5 5 z = a sec = 2a cos 4. Gravitational attraction (directed along z axis, by symmetry): (G/ ) cos dV = G R 2 /4 0 0 R a sec 2 /2 /4 0 2a cos 2 y x cos sin d d d. (here = 1) cos sin d d d + Must split into: G( 0 cos sin d d d). 0 First term: Inner: Middle: /4 0 a sec 0 cos sin d = a sin ; /4 0 a sin d = -a cos 2a cos 0 = a(1 - 2 2 ); Outer: 2a(1 - 2 4 2 2 ). Second term: Inner: Middle: /2 /4 cos sin d = 2a cos2 sin ; /2 /4 2 = 3a 2 2a cos2 sin d = - 3 a cos3 2 3 ). =a 2 6 ; Outer: 2a 2 6 . Final answer: F = 2Ga(1 - 2 ...
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