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## Solutions to Week 11 questions (the Regular and Physics section problems are almost identical)

 1) The boundary of the triangle consists of segment A where z = 0, x + y = 1; segment B, where x = 0, y + z = 1; and segment C where y = 0 and z + x = 1. Traversed clockwise, these segments are parametrized for time 0 (1-t,t,0), for B, t->(0,1-t,t), for C, t->(u,0,1-t). The interior of the triangle is parameterized by (u,v) with 0 (u,v,1-u-v). For this parameterization, XuxXv is the vector is (1,1,1). The flux integral of curl(F) over the interior of the triangle uses the normal in the direction of (1,1,1). Using the parametrization given above, this is the integral of curl(F)|(x=u,y=v,z=1-u-v).(1,1,1) over the triangle in the (u,v) plane, where 0 a) The line integral over A is 1/2, over B it is 1/2, and over C it is also 1/2. The flux integral, 3/2, of curl(F) is that of curl(F).(1, 1, 1) = 3 over the triangle. The latter has area 1/2. b) The line integral over A is 1/6, over B it is 0, and over C it is -1/6. The flux integral of curl(F) over the interior of the triangle is 0 as curl(F) = 0. c) The line integral over A is -1/6, over B it is -1/2, and over C it is 0. The flux integral, -2/3 of curl(F) is that of curl(F)|(x=u,y=v,z=1-u-v).(1, 1, 1) = -(1 + u) over the region in the (u,v)-plane where u is between 0 and 1 and v is between 0 and 1-u. 2) a) The flux is 1 through the faces where either x, y or z has constant value 1, and it is -1 through the faces, where either x,y or z has constant value 0. The volume integral for the divergence theorem gives zero as div(F) = 0. b) The flux is 0 through the faces where y and z is constant and through the face where x has the constant value 0. The flux is 1 through the remaining face. For the volume integral part of the divergence theorem, div(F) = 1, so the volume integral of div(F) gives the volume of the cube, which is 1. 3) Any plane with normal vector perpendicular to curl(F) will have this property, for example the plane where x-y = 0. 4) These are both exercises in the product rule for taking derivatives. 5) The divergence theorem says that the integral of div(F) over the interior of the ball is that of F.n over the boundary sphere, where n is the unit normal vector. Now, F.n is equal to |F| cos (q), where q is the angle between F and the normal. Since |cos(q)|< 1, the integrand for the surface integral is no larger than |F|<1 and no smaller than -|F|>-1. Thus, the integral of F.n over the surface is no larger than 1 (area) and no smaller than -1 times the area. As the area of the sphere is 4 , the claim follows. 6) According to Stokes' theorem, the integral here is the integral of the path integral of F around the boundary. Parametrize the boundary by j with 0 via ( (cos(t),sin(t) ). Then, this path integral is the integral between 0 and 2 of F|(x-cos(t),y-sin(t),0).(-sin(t), cos(t), 0). As (-sin(t) cos(j) is a unit vector, this integrand is no larger than |F| <1 and no smaller than -|F| >-1, so the integral is no larger than 2 and no smaller than - 2 . 7) The iterated integral is the same as = 1/6. As div(F) = 1, the flux integral through the boundary should also give this same answer. The flux of F through the boundaries where either x = 0, y = 0 or z = 0 is zero. That of F through the boundary where x+y+z = 1 is that of u=F|(x=u,y=v,z=1-u-v).(1,1,1) over the region in the (u,v) plane where u is between 0 and 1 and v is between 0 and 1-u. This integral is also 1/6. 8) According to Stokes' theorem, the integral is the same as the absolute value of the integral of curl(F).n over the interior of the disk D. Since (cos(q),0,sin(q)) is normal to this disk, and curl(F) = (1,0,0), this last integral has absolute value |cos(q)|. Hence, it is maximized, where q=0.

 Last update, 12/6/2001