This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 18.02A IAP 2010: Solutions Problem Set 8 Part II Problem 1.a) Using @ @x x 3 = 1 3 + 3 x 2 4 and similarly for y and z , we compute that r Â¡ F = GM 3 3 + 3( x 2 + y 2 + z 2 ) 4 = 0 except at (0 ; ; 0), where = 0, and the divergence is unde ned. b) Let S a be the sphere of radius a with normal oriented outward, so n = ^ = x i + y j + z k a . Then ZZ S a F Â¡ n dS = GM Z 2 Z a 2 a 4 a 2 sin d d = 2 GM ( cos ) = 4 GM; independent of the radius a . c) Let S be an arbitrary closed surface containing the origin (so it also encloses a sphere of radius a centered at the origin, for some possibly small a ), with outward oriented normal. Let D be the region between S and S a , so @D = S S a , taking into account our orientation convention on S a . By the divergence theorem then, ZZ S F Â¡ n dS ZZ S a F Â¡ n dS = ZZZ D dV = 0 so ZZ S F Â¡ n dS = ZZ S a F Â¡ n dS = 4 GM: Problem 2.a) From the description, the bouancy force on the submerged body will be F b = ZZ S g ( z z ) n dS where z is the free surface of the uid, and S is the boundary of the submerged body. This is a vector quantity, with components F b = F x i + F y j + F z k where F x = ZZ S g ( z z ) i Â¡ n dS 1 and similarly for F y and F z , with i replaced by j and k , respectively. Note that F x is equivalent to the ux of a vector eld G x = g ( z z ) i , whose divergence vanishes and so F x (and similarly...
View Full Document
- Winter '10
- Trigraph, Cxy, CYZ