# Use the force field to find the work done on a particle.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

The work done by the force field in moving the particle along a path is a circulation, or line integral, of this force field around the path. The circulation is defined as

`W = int_C vecF* dvecs` .

The expression under the integral is the scalar product of the force field and the vector `dvecs` , which is tangent to the line.

The given force field has only a y-component, so the line integral of this force field on the horizontal pieces of the given path (`C_1`  and `C_3`  ) will be zero. (The scalar product of two perpendicular vectors is zero.)

So the work of the given force field will be

`W = int_ (C_2) vecF*dvecs + int_(C_4) vecF*dvecs` .

Since `C_2`

is vertical and is traversed upward, for `C_2`

`dvecs = dvecy` . Since `C_4` is vertical and is traversed downward, for `C_4` `dvecs =-dvecy` .

The scalar product of the unit vector `vecj`

and `dvecy`

is `vecj*dvecy = dy`

because they are parallel.

Therefore, the integral for work above becomes

`W = int_(C_2) (-(x + 2y^2))dy + int_(C_4) (-(x + 2y^2))(-dy)`

To evaluate these integrals, consider their boundaries. The curve `C_2`

is the segment of straight vertical line with the equation x =3, bounded by the y-values y = -1 and y = 1. The curve `C_4`

is the segment of straight vertical line with the equation x = 1, bounded by the y-values y = -1 and y = 1. So, after the integrals with respect to y are taken, x = 3 can be plugged in the first resultant function, and x = 1 in the second resultant function:

`W = -xy|_(3, -1) ^(((3, 1))) -2y^3/3 |_(-1) ^ 1 + xy |_(1, -1) ^(((1, 1))) + 2y^3/3 |_(-1) ^ 1 = -3*2 +1*2 = -4`

(In the first and third expression, (3, 1) and (1, 1) should be in parentheses but the math editor is refusing to display them.)

The work performed by the given force field in moving a particle around the given path is -4.

This result can also be obtained by using Green's theorem. It states that the circulation of the force field around a closed path can be related to the area enclosed by the path:

`int_C (Pdx + Qdy) = int int ((dQ)/(dx) - (dP)/(dy)) dA` .

Here, `P = F_x`

and `Q = F_y`

For the given force field, P = 0 and `Q = -(x + 2y^2)`

so `(dQ)/(dx) = -1`

This means the surface integral equals negative area of the region enclosed by the given path. This region is the square with the side 2, so its area is 4. Thus, the surface integral equals -4, which is consistent with the result for circulation obtained previously.