where C is the top half of a circle (centered at the origin) of radius
2 along with the straight line segment from (-2,0) to (2,0).
C is a simple closed curve and the partials are
fine so Green's applies, producing 
.
Both the region and the integrand look easier to handle in polar coordinates,
so we convert to 
.
This integral is straightforward and works out to 4
.
2. Compute the curl of the vector field F(x,y,z) = i
- 2x j + y k.
We set up the usual determinant 
, which works out to (1-0)i + (0-0)j + (-2-0)k = i
- 2k.
3. Compute the divergence of the vector field
F(x,y,z) = xey i - ze-y
j + y ln z k.
We work out the partials with respect to x, y,
and z, and add them, to get ey + ze-y
+ y/z, a function found nowhere in nature.