F=(ycosx,sinx)f=y(1+4x^2)^.5pa... from c: A=(0,0) to B=(1,1) along y=x^2scalar line integral fds =?

Consider the field F: (x,y)=(ycosx,sinx)


and the scalar field f: f(x,y)=y(1+4x^2)^.5


and consider the path from c: A=(0,0) to B=(1,1) along y=x^2


--Determine the scalar line integral of fds


--Determine the vector line integral of Fds





Q: is F a gradient field?

F=(ycosx,sinx)f=y(1+4x^2)^.5pa... from c: A=(0,0) to B=(1,1) along y=x^2scalar line integral fds =?
Parametrise the path A to B by





r(t) = (t, t^2) for t = 0 to 1,





so that





r'(t) = (1, 2t).





The scalar line integral of f along the path is given by





∫ f.ds ∫ f.|r'(t)|.dt = ∫ t^2.(1 + 4t^2)^(1/2).(1 + 4t^2)^(1/2).dt





= ∫ t^2 + 4t^4.dt,





where the integral is taken from t = 0 to 1. Evaluate this to get your answer.





For the vector line integral, first observe that u(x, y) = y.sin x has F as its gradient (it is indeed a gradient field). Then, along our parmetrised line,





d[u(r(t))]/dt = Grad(u(r(t))•r'(t) = F(u(t))•r'(t),








and so the vector line integral of F is given by





∫ F•dr = ∫ F(u(t))•r'(t).dt = ∫ [d[u(r(t))]/dt ].dt = ∫ d[u(r(t))] = u(r(1)) - u(r(0))





= (1^2).sin 1 - (0^2).sin0 = sin 1.