1.  (10 points)  Find the area of the region bounded by the line y=x and the curve

	r(t) = x(t) i + y(t) j = (t^3, 2 t^2),  0 ² t ² 2
a)  Express, but do not evaluate, the area as a specific line integral:

The idea here is to use Green's theorem to write the area as a line integral of x dy (or -y dx):
> int(y, y=0..8) + int(t^3 * 4*t, t=2..0);

                                 32/5

b)  Express, but do not evaluate, the area as a specific integral in the changed variables  
u = x, v = y^3/x^2.  (Hint:  if x=y, then v = u.)

Don't forget the Jacobian:  Since x = u, y = (v u^2)^(1/3), 
     dx dy = |det [[1,0],[(2/3)(v/u)^(1/3), (1/3)(u/v)^(2/3)]| du dv

we integrate in u from 0 to 8 and in v from u to 8

> int((1/3)*(u/v)^(2/3), v=u..8);

                                           2
                             2/3  2/3     u
                       1/8 64    u    - -------
                                          3 1/3
                                        (u )

> int(", u=0..8);

                           2/3  2/3          2/3
                     3/5 64    8    - 1/2 512

> simplify(");

                                 32/5

c)  Evaluate the area:

Actually, it is also a "kindergarten integral":

> int(2 * x^(2/3) - x, x=0..8);

                                  2/3
                            48/5 8    - 32

> simplify(");

                                 32/5

2.  (10 points)  Let F = x^3 i + (x + z) j + (4 y - z) k.  and consider the solid ½ bounded by z=2 and
the cone f = ¹/6.  (The angle f is the colatitude.)  Integrate F.n over the surface.
The best way is to use the divergence theorem and cylindrical or spherical coordinates.  Let us choose cylindrical.  
a)
The divergence of F is
3 x^2 - 1 = 3 r^2 cos(theta)^2 - 1, and we integrate in :  theta from 0 to 2 pi; r from 0 to z/2, and z from 0 to 2.  The volume element is r d theta dr dz .

b)
> int(3 * r^3*cos(t)^2 - r, t=0..2*Pi);

                                 3
                           3 Pi r  - 2 r Pi

> int(", r=0..z/2);

                                 4        2
                        3/64 Pi z  - 1/4 z  Pi

> int(", z=0..2);

                                 11
                               - -- Pi
                                 30

3.  (Essentially a graphical problem).
4.  (10 points)  The curve C

	(x(t), y(t), z(t)) =  (2 cos(2t), 2 sin(2t), 4 cos(2t))

is an ellipse, and we consider  f(x,y) = x - 2 y + 3 z  on C.

a)  To maximize f on C we could use Lagrange, or we could substitute and use the chain rule.
With the latter we have 14 cos(2t) - 4 sin(2t), which is maximized when tan(2 t) = -2/7:
> subs(t=-(1/2)*arctan(2/7), 14*cos(2*t)-4*sin(2*t));

              14 cos(-arctan(2/7)) - 4 sin(-arctan(2/7))

> simplify(");

                                   1/2
                               2 53

c)  Let's do c next, since we can see that the plane of the ellipse is z= 2 x, or -2x + z = 0.  This tells us that the binormal is b=(1/sqrt(5)) (-2,0,1), which helps with b):
b)  t = [- sin(2t),  cos(2 t), -2 sin(2 t)] /(sqrt(1+4 sin(2 t)^2)
as remarked above, b = (1/sqrt(5)) (-2,0,1).  hence
n = b X t =  [-cos(2 t),  -5 sin(2 t), -2cos(2 t) /sqrt(5 + 20*cos(2 t)^2)  .
> 
d)  The maximal curvature will occur where the ellipse is n the long axis of the ellipse, i.e., when t=0, pi, etc.  We differentiate the tangent vector by the arclength by recalling that df/ds = df/dt/ (ds/dt):
> diff([- sin(2*t),  cos(2*t), -2 *sin(2*t)] /sqrt(1+4*sin(2*t)^2), t)/(4*sqrt(1+4*sin(2*t)^2));

    /[-2 cos(2 t), -2 sin(2 t), -4 cos(2 t)]
1/4 |---------------------------------------
    |                        2 1/2
    \         (1 + 4 sin(2 t) )

         [-sin(2 t), cos(2 t), -2 sin(2 t)] sin(2 t) cos(2 t)\   /
     - 8 ----------------------------------------------------|  /
                                        2 3/2                | /
                         (1 + 4 sin(2 t) )                   /

                   2 1/2
    (1 + 4 sin(2 t) )

When we substitute t=0, we have
> [-2,0,-4]
so the curvature is
> sqrt((-2)^2+(-4)^2);

                                   1/2
                                2 5

> 
5.  (10 points)  Evaluate the integral of (3x^2 y)i + x^3 j + 2 k over the curve of Problem 4.

We note that this vector field is the gradient of x^3 y + 2 z.   Therefore we can simply evaluate this function at the endpoints.
a)  In this case the end points are (2,0,4) and (-2,0,-4), so the integral is -16.
b)  In this case the curve is closed, so the integral is 0.
6.  (10 points)  The surface ½ defined by

	5x^2 + y^2 + z^2 - 4x z = 4, z > 2x

is half of an ellipsoid. (Possibly helpful:  x^2 + y^2 + (z-2x)^2 = 4.)  Let n be the upward normal vector to the ellipsoid and define the vector field:

	F =  (3 y + 3 x^2y)i + (x+x^3) j + 2 k  .
Evaluate the integral of curl(F).n over this surface.
We are meant to use Stokes's theorem, preferably twice.  The boundary of this ellipsoid is the ellipse of Problems 4-5.  As established in Problem 5, 3 x^2 y i + x^3 j + 2 k is a gradient, so the integral of this piece around a closed ellipse iscontributes 0.  The curl of F is in fact just (0,0,-2).  By Stokes, we can integrate this over any surface bounded by the ellipse; so let us shoose to integrate it over the part of the plane z=2x contained within the ellipse.  The normal is (-2,0,1)/sqrt(5), so we the integral is -2/sqrt(5) times the area of the ellipse.  Since the area of the ellipse is the integral of |N| dx dy over the circle of radius 2, where N = (-2,0,1), we get:
        -2 * pi (2)^2 = -8 pi