Let F=xy2i^+(y+x)j^. Integrate (∇×F)⋅k^ over the region in the first quadrant bounded by the curves y=x2 and y=x using Green’s theorem.
Technique
(∇×F)⋅k^=Qx−Py=1−2xy; integrate over R with x2≤y≤x, 0≤x≤1.
Solution
Step 1 — The integrand (∇×F)⋅k^
With F=(P,Q)=(xy2,y+x) (a planar field),
(∇×F)⋅k^=∂x∂Q−∂y∂P=∂x∂(y+x)−∂y∂(xy2)=1−2xy.
By Green’s theorem this double integral equals the boundary line integral ∮∂R(Pdx+Qdy); we evaluate the double integral directly (both are computed below).
Step 2 — Describe the region
The curves y=x2 and y=x meet where x2=x, i.e. x=0 and x=1. On [0,1] the line lies above the parabola (x≥x2), so