Let S be the part of the surface z=x^2+y^2 between the planes z=1 and z=2.

Let C be the boundary of S that has positive z-component. Let F be vector field F(x,y,z)=. Evaluate integral F*dr with bound C

(a) directly as line integral
(b) as a double integral, by using Stoke's Theorem

(a) Note that the outer circle x^2 + y^2 = 2 has clockwise orientation, while the inner circle x^2 + y^2 = 1 has counterclockwise orientation.

Parameterizing as follows:
x^2 + y^2 = 2 ==> x = √2 cos t, y = √2 sin t, z = 2 with t in [0, 2π].
x^2 + y^2 = 1 ==> x = cos t, y = sin t, z = 1 with t in [0, 2π] with opposite orientation

So, ∫c F · dr
= ∫(t = 0 to 2π) <√2 sin t, -√2 cos t, 2> · <-√2 sin t, √2 cos t, 0> dt
- ∫(t = 0 to 2π) · <-sin t, cos t, 0> dt

= ∫(t = 0 to 2π) -2 dt - ∫(t = 0 to 2π) -1 dt
= -2π.
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(b) Note that curl F = <0, 0, -2>.

Moreover, the region of integration (via cartesian coordinates) is
1 < x^2 + y^2 < 2.

So by Stokes' Theorem, the line integral equals
∫∫s curl F · dS
= ∫∫ <0, 0, -2> · <-z_x, -z_y, 1> dA
= ∫∫ -2 dA
= -2 * (Area inside 1 < x^2 + y^2 < 2)
= -2 * (2π - π)
= -2π.

I hope this helps!