If $\displaystyle F = (x+4)i +xyj + (xz-y)k$, evaluate $\displaystyle \int c F.dr$ along the following paths C:
the straight line joining (0,0,0) to (2,1,1)
I don't really understand what to do and what the limits are.
Thanks Adam
Ok
Im not getting the same answer as my book
(1-t)(0,0,0) + t(0,0,1) = (0,0,t)
and (1-t)(0,1,1) + t(2,1,1) = (2t,1,1)
So for the first one
r' = (0,0,1)
F.r' = 0
So the integral of that is 0
Then for the second one
r' = (2,0,0)
F.r' = 4t + 8
so the integral between 1 and 0 = 10
But the answer in my book is 38/3