ENGI 5432 Advanced Calculus

    Faculty of Engineering and Applied Science
    2009 Winter

    Term Test 2   -   Solutions

    1. For the vector field defined by   F = < yz, zx, xy >,

      1. Find the divergence of   vector F   vector F


        div F = 0

        div F = 0


      2. Hence find the total flux   scalar capital Phi   of   vector F   through the sphere of radius 1, centre (0, 0, 0).

        Using Gauss’ divergence theorem,
        Phi = 0

        Phi = 0

        [Note that the details of the sphere are irrelevant - for this source-free vector field, the net flux through any simple closed surface will be zero!]


    1. In the x-y plane a vector field is defined by F = < 3x^2 y, x^3 - y + 2x >.

      1. Use Green’s theorem to find the work done by   vector F   in one circuit around the unit square (vertices (0, 0), (1, 0), (1, 1) and (0, 1)).

        [image of unit square] Green’s theorem states that, in any simply connected domain,
        line integral on C of F dot dr = 
integral over enclosed area of normal component of curl F
        where   C   is a simple closed path that encloses the region   D.
        z-component of curl F = 2
        W = 2A
        But   D   is a unit square, whose area is   A = 1.   Therefore

        W = 2


      2. Is the vector field   vector F   conservative?   Why or why not?

        curl F is not zero

          NO,   F   is not conservative.  


    1. The velocity field of a fluid is   v = 2 kHat.   Find the flux   Q   of the fluid through the hemisphere   x^2 + y^2 + z^2 = 4 ;   z >= 0   in a direction outward from its centre.
      [Hint:   use a spherical polar coordinate grid   x = 2 sin theta cos phi ,
y = 2 sin theta sin phi, 
z = 2 cos theta,   with   0 < theta < pi/2 ,  0 < phi < 2 pi


      [image of hemisphere] vector r = ...
      dr/d_theta = ...
      dr/d_phi = ...
      N = ± dr/d_theta cross dr/d_phi
      N = ...
      But, everywhere on the hemisphere,   cos and sin (theta and phi)   are all positive.
      For an outward pointing normal vector (pointing away from the origin), we must select the '+' sign.
      N = ...
      v dot N = 8 sin theta cos theta
      Q = Integral_S v dot dS
      Q = ...
      Therefore the flux is

      Q = 8 pi


    1.   BONUS QUESTION

      Find the potential function   phi   for   F = < y cos x, (sin x + z sin y), (z - cos y) >   such that   phi = 1   everywhere along the y-axis.


      F = grad phi
      matching partial d_phi/dx
      matching partial d_phi/dy
      revised phi
      matching partial d_phi/dz
      phi = y sin x  -  z cos y  + z^2 / 2 + C
      But   phi = 1   everywhere along the y-axis
      C = 1
      Therefore the required potential function is

      phi = y sin x  -  z cos y  + z^2 / 2 + 1


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    Created 2009 03 03 and most recently modified 2009 03 04 by Dr. G.H. George