This is a closed path
check first if field is potential in which case integral is zero
F = z^2 i + 2xy j + 3y^2k
del cross F = 0 ????
nope no way
so first line =
3 i + 0 j + 0 k
F dot line = Fx *3 = 0*3 = 0
second line
in x from 3 to 3 = 0
in y from 0 to 3 = 3
in z from 0 to 1 = 1
Fy = 2 x y = 6 y
y work = 6 y dy from 0 to 3 = 6*9/2 = 27
z work = 3 y^2 from z =0 to z = 1
but y = 3 z so 27 z^2 from 0 to 1
27 z^3/3 0 to 1 = 9
so total work in line 2 = 27+9 = 36
so far 27 + 36 Joules
now do lines three and four
and get the total for all four
and I am thinking you are in MITx as well :)
A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field
F(x, y, z) = z2i + 2xyj + 3y2k.
Find the work done.
2 answers
I'm very confused on this?
1 answer