high level math

Along the path (0,0)->(2,0)
F=xy i + y^2 j=0
∫F(r)．dr=0

Along the path (2,0)->(0,3)
i.e. along y=3-3x/2
Parametrizing the path and the field:
r = t i+(3-3t/2) j
dr/dt=i -3/2 j
F=xy i + y^2 j=t(3-3t/2) i + (3-3t/2)^2 j

∫F(r)．dr=∫(2->0)[t(3-3t/2) i + (3-3t/2)^2 j]．[i -3/2 j] dt
=∫(2->0)[t(3-3t/2) - 3(3-3t/2)^2 /2]dt
=∫(2->0)[t(3-3t/2) - 3(3-3t/2)^2 /2]dt
=∫(2->0)[3t-3t^2 /2 - 3(9-18t/2 +9t^2 /4)/2]dt
=1/8 ∫(2->0)[24t-12t^2 -108+108t-27t^2]dt
=1/8 ∫(2->0)[-39t^2 +132t -108]dt
=-3/8 ∫(2->0)[13t^2-44t +36]dt
=-3/8 [13t^3/3 -22t^2 +36t](2->0)
=3/8 [13(2^3)/3 -22(2^2) +36(2)]=13-33+27=7


Along the path (0,3) --> (0,0)

∫F(r)．dr=∫(3->0) (y^2 j)．(jdz)
=∫(3->0) y^2 dy
=[y^3 / 3](3->0)
=-9


∫F(r)．dr=7-9=-2
C