Math B6C – Exam 5 – SOLUTIONS 

*  Fall 2002 *

1.  Calculate the line integral, , with   ,  over the elliptical path g  parametrized by        from    t = 0   to   t = p.

Since   and   ,  we get:

2.   Calculate the line integral, , with   ,  over the parabolic path g  parametrized by     from   t = 0   to   t = 1.

  from   t = 0   to   t = 1.

Since   and  ,  we get:

 3.   Calculate   with   ,  where   g(x, y) = 5x3 – 2xy2,   over the path  g   parametrized by       from   t = 0   to   t = 1.

 Since ,    we have:

 

 4.   Suppose that    with  g(0,0) = 0.      Then   g(x, y) = ?

 

5.  Calculate      with    ,   over the path g  from the point (1, p) to  (2, p/2)  along a straight line.    

 

6.    Suppose that    with  g(0,0) = 0.     Then   g(x, y) = ?

 

 

 

7.   Calculate    with    ,   over the path g  from the point ( 3, 1 ) to ( 5, 2 ) along the line segment connecting these two points.

 

8.   Calculate   with    ,   over the path g  from the point (p, –1)  to (0, 1)  along the curve    y = cos x.

 

 9.  If    and  g  is the boundary of the square with
      vertices  (1,1), (3,1), (3,3), (1,3),  then  

By Green’s theorem we have that  

 

10.  Suppose    and region R has an area of  7  and has boundary g  that is a simple closed curve.        Then  

By Green’s theorem,

.

 

11.  Suppose , where   g(x,y,z) =  x2 cos y – sin(yz).    If  g  is the ellipse parametrized by  

 

F is a gradient field, and so over any closed path such as g.

 12.  If  g  is the circle  x2 + y2 = 4,  then  

This integral is the circulation of about the given circle of radius 2.

By Green’s theorem,

. 

Changing to poplar coordinates, we get 

13.  If   and  g  is the unit circle centered at the origin.  If R is the circular interior of  g,  then          

              [ Hint:  this time it’s easier to compute the line integral! ]

 By Green’s theorem,  .

We will use the unit-circle parametrization,  ,   from t = 0  to  t = 2p. 

14.  Compute the (upward) flux of the field  through the portion of the paraboloid surface  z = 1 – x2 – y2   that is above (or below) the rectangle in the xy-plane with corners at (0,0), (1,0), (1,1), (0,1).

 

The flux is given by the integral     
Let  G(x,y,z) = 1 – x2 – y2 – z  , so the S is the surface implicitly described by  G(x,y,z) = 0.    Notice that the upward normal to this surface is the negative of the gradient of G, not the gradient of G.   

Then,

 

15.  Compute the divergence of the vector field  .

16.  Compute the outward flux of through the surface of the cube defined by   .


Invoking the divergence theorem,

 

17.   Compute the curl of the vector field .

18.   Compute the circulation of around the rectangular path g  from (0,0,0) to (1,0,0) to (1,1,1) to (0,1,1) and then back to (0,0,0).             

By Stoke’s theorem,  ,  S being the interior of the given square.    Since S is on the plane  y = z,  i.e. the plane   0x + 1y –1z = 0,  and so has normal vector (0,1,­–1).   Thus the unit normal to the square is , and so  .   Using the results from problem #17,  we have

, and since on this rectangular surface we have  y = z,  this becomes    and so

 19.   Suppose  g(x,y,z) = xy2z3.   Then  

   The curl of a gradient field is zero:     . 

 20.  Compute the circulation of the field (featured in #19),  around the rectangular path g,   from  (0,0,0)  to (1,0,0) to (1,1,1) to (0,1,1) and then back to (0,0,0).     

The circulation around any closed curve is zero, for any gradient field:  .