1. Compute the surface integral  where F(x,y,z) = yi - xj - 3k and S is the surface of the paraboloid z = x² + y² (with upward orientation) within the cylinder x² + y² = 1.

I. We parametrize the surface in our usual way, getting
x(u,v) = u
y(u,v) = v
z(u,v) = u²+v²
so r(u,v) = <u,v,u²+v²>, for values of u and v for which u²+v² is less than or equal to 1 (to stay within that cylinder).

II. We find F(r(u,v)) = <v, -u, -3>.

III. We find r_u=<1,0,2u> and r_v=<0,1,2v>, and then cross them to get r_u # r_v = <-2u,-2v,1>.

IV. We put the last two pieces together to form the integral .

V. We dot the vectors in the integrand and convert to polar coordinates to get  which works out to -3.
 
 

2. Use Stokes' Theorem to compute  where S is the top half of a sphere of radius 3 centered at the origin with upward orientation and F(x,y,z) = xi + yj + zk [Note that it should come out to zero, since curl F = 0, but show that Stokes' gives zero too].
 

By Stokes' Theorem, this surface integral is the same as , where C is the circle which forms the boundary of the hemisphere, i.e. the circle of radius 3 centered at the origin in the xy plane. So we compute this line integral:

I. We parametrize this circle (noting that each of its points has a z coordinate of 0 since it lies in the xy plane) as
x(t) = 3cos t
y(t) = 3sin t
z(t) = 0
so r(t) = <3cos t, 3sin t, 0> for 0  t  2

II. We find F(r(t)) = <3cos t, 3sin t, 0>

III. We compute r'(t) = <-3sin t, 3cos t, 0>

IV. We set up the integral .

V. We compute the dot product to make the integrand -9sin t cos t + 9sin t cos t, which is 0, and of course integrating this from 0 to 2 gives 0.