Not quite
Green's Thm --- Line Integral = integral(dg/dx-df/dy) over the region enclosed by the C
you computed dg/dx-df/dy correctly = 1
so the line integral is the area of the triangle then which is 1/2
Here is the region of interest. In order to set up
then we would need the following
.
Since the inner integral involves then the curve to cuve in in the direction and in your case and hence the inside integral . Then outer integral is point to point and since the outer involves then we are moving in the direction and from the picture of the region it and hence the outer integral of
Hope that helps.