(950, #21) Evaluate

where F(x,y,z) = < 0, y, -z > and S consists of the paraboloid y = x² + z², 0 <= y <= 1, and the disk x² + z² <= 1, y = 1.

Solution: The surface has two distinct pieces, the paraboloid and the disk at the end:

On the paraboloid:

x = u cos(t), y = u², z = u sin(t), u = 0..1, v = 0..2 Pi

and

Since the second component of this normal is -u, the normal to the parabolic piece will be pointing to the left.

Thus, if P is the parabolic part then

On the disk we use the parametrization

x =u sin(t), y = 1, z = u cos(t), u = 0 ..1, t = 0..2*Pi.

Then

This normal points to the right of the disk. Thus the selected parametric equations yield a system of normals pointing out of the surface.

If we let D be the disk we have

Consequently, the sum of the two integrals is 0.