Evaluate line integral of vector field

Question

F (x, y, z) = y sin(x2) i + xyz j + (x + y + z) k along line segment from the point (0, 0, 0) to point (1, 2, 3).

Answer 1

Parameterize the path from (x,y,z) = (0,0,0) to (x,y,z) = (1,2,3)

as [x=t, y=2t, z=3t] for (0 ≤ t ≤ 1).

From F(x,y,z) = y sin (x²)i + (xyz)j + (x+y+z)k, translate

to F(t,2t,3t) = 2t sin (t²)i + (t•2t•3t)j + (t+2t+3t)k or

2t sin (t²)i + (6t³)j + (6t)k.

From ∫_{[from A=(0,0,0) to B=(1,2,3)]} F•ds equals

∫_{[from t=0 to t=1]}[F_xdt/dt + F_yd(2t)/dt + F_zd(3t)/dt]dt,

write ∫_{[from t=0 to t=1]}[2t sin (t²)•1 + (6t³)•2 + (6t)•3]dt.

Simplify to ∫_{[from t=0 to t=1]}[2t sin(t²) + 12t³ + 18t]dt.

Integration gives [-cos (t²) + 3t⁴ + 9t²|_{[from t=0 to t=1]}]

or 12 − cos (1 Radian) − 0 + cos (0 Radian) or

13 − cos (1 Radian) equal to 12.45969769.