Line Integral?

Question

∫c(x+y+z)dx+xdy-yzdz, where c is the line segment from (1,2,1) to (2,1,0)

Answer 1

Parameterize the line segment by

(x,y,z) = (1,2,1) + t*[(2,1,0) - (1,2,1)] = (1+t,2-t,1-t).

So x=1+t, y=2-t, z= 1-t.

So dx=dt, dy=-dt, dz=-dt.

Substitute these into the integral to obtain the integral with respect to t between t=0 and t=1 of

[(1+t) + (2-t) + (1-t)]*1 + (1+t)*(-1) -(2-t)*((1-t)*(-1) =

t^2 - 5t + 7.

This integrates to 1/3 -5/2 +7 = 29/6.

Answer 2

break it up:

(x+y+z) is integrated with respect to x = x^2/2 + yx + zx
(x) is integrated with respect to y = xy
(yz) is integrated with respect to z = yz^2/2

F(x,y,z) = [ x^2/2 + 2xy + yz^2/2 + zx ]

now evaluate over the distance between A(1,2,1) and B(2,1,0)