If `vec(F) = y hat(i) + x hat(j)` then find out the work done in moving the particle from position (2,3) to (5,6)

Explanation

dw = `vec(F).d vec(s)`
dw = `(y hat(i) + x hat(j)).(dx hat(i) + dy hat(j) )`
dw = ydx + xdy
Now ydx + xdy = d(xy) (perfect differential equation )
`rArr` dw = d(xy)
for total work done we integrate both side
`int " dw " = int` d(xy)
Put xy = k
then at(2,3) `k_(i) = 2xx3 = 6 `
at`(5,6) k_(f) = 5 xx6 = 30`
then w = `int_(6)^(30) dk = [k]_(6)^(30)`
`rArr " " w =` (30 - 6) = 24 joule