
Work: Vector Integration
The force function $\displaystyle \bold{F} = \bold{\hat{x}}kx  \bold{\hat{y}}ky$ is using Cartesian coordinates.
Find the work done from (1,1) to (4,4) using the following path:
$\displaystyle (1,1) \to (1,4) \to (4,4)$
So, what I did was:
$\displaystyle \int^{4,4}_{1,1} \bold{F}\cdot ~d\bold{r}$
$\displaystyle = \int^{4,1}_{1,1} \bold{F}\cdot ~d\bold{r} + \int^{4,4}_{4,1} \bold{F}\cdot ~d\bold{r}$
$\displaystyle = \int^{4,1}_{1,1} (kx ~dx  ky ~dy) + \int^{4,4}_{4,1} (kx ~dx  ky ~dy)$
$\displaystyle = \int^4_1 (kx ~dx) + \int^1_1 (ky ~dy) + \int^4_4 (kx ~dx) + \int^4_1 (ky ~dy)$
$\displaystyle = k\left(\int^4_1 x ~dx + \int^1_1 y ~dy + \int^4_4 x ~dx + \int^4_1 y ~dy\right)$
$\displaystyle = k\left[\left(8  \frac{1}{2}\right) + \left(8  \frac{1}{2}\right)\right]$
$\displaystyle = k(15) = 15k$
Is this right?

It seems to me that if you are heading in the xdirection the work done will only depend on the x component of the force, and if you are heading in the y direction the work done will only depend on the ycomponent of the force, so you can add $\displaystyle \int _1 ^4 f(y) dy + \int _1 ^4 f(x) dx $ where f(y) is the ycomponent of force and f(x) is the xcomponent of force. Your answer looks right to me.

i agree. your answer is correct. i've never seen that method before, though. it seems to make sense

I've never seen this method either... But it's the way me and my friends interpreted it given this example:
The force exerted on a body is $\displaystyle \bold{F} = \bold{\hat{x}}y + \bold{\hat{y}}x$. The problem is to calculate the work done going from the origin to the point (1,1):
$\displaystyle W = \int^{1,1}_{0,0} \bold{F} \cdot ~d\bold{r} = \int^{1,1}_{0,0} (y ~dx + x ~dy)$
Separating the integrals, we obtain:
$\displaystyle W = \int^1_0 y ~dx + \int^1_0 x ~dy$
The first integral cannot be evaluated until we specify the values of x as y ranges from 0 to 1. Likewise, the second integral requires x as a function of y. Consider the first path:
$\displaystyle (0,0) \to (1,0) \to (1,1)$
Then:
$\displaystyle W = \int^1_0 0 ~dx + \int^1_0 1 ~dy = 1$
since y = 0 along the first segment of the path and x = 1 along the second. If we select the path:
$\displaystyle 0,0 \to (0,1) \to (1,1)$
then the integral gives $\displaystyle W = 1$. For this force the work done depends on the choice of path.
That's pretty much all we had to go by. No professor, no class, just us and a book...
We did have this relation:
$\displaystyle W = \int \bold{F}\cdot ~d\bold{r} = \int F_x(x,y,z) ~dx + \int F_y(x,y,z) ~dy + \int F_z(x,y,z) ~dz$
Using that, the separation of the integrals follows.