Is the work integral decomposable?

Question

Work is defined as $W = \int_{\gamma} \vec F \cdot d\vec l$ which I think means $W = \int (F_x, F_y, F_z) \cdot (dx, dy, dz)$. So by the linearity of the integral, could we always decompose work into the integrals in each of the directions $x,y,z$? As in, does $W = \int_{x_0}^{x_1} F_xdx + \int_{y_0}^{y_1} F_ydy + \int_{z_0}^{z_1} F_zdz$? Because that seems easier to compute (at least sometimes).

Answer 1

I can't see how it differs from the usual method. I suppose you mean the first integral depends on $x$ only, the second one on $y$ only and the third one on $z$ only, so that simplifies the calculation. But this is not true. In fact, $F_x$ is not a function of $x$, it is not even a function of $x$, $y$, $z$, but a function of time $t$. So at last, you need to write everything it in terms of the parameter $t$. So that's the same way it's usually done, exactly that you write it explicitly as the sum of three integrals.

Besides, there are some problems in your using such a notation. For example, if the particle moves along a closed path, do you write $\int_{x_0}^{x_0}$ etc.? That will suggest that the work done is always zero.

Answer 2

You can correctly write $$W = \int_\gamma F_x\,dx + \int_\gamma F_y\,dy + \int_\gamma F_z\,dz.$$

But note that this is still tracing the path $\gamma$, so $\int_\gamma F_x\, dx$ is not a univariate integration in the general case. If $F_x$ is a function of $x$ only, then you can write $\int_\gamma F_x\, dx = \int_{x_0}^{x_1} F_x\, dx,$ but that's a very restrictive condition. It requires at a very minimum that the force is conservative, so that how much work is done is a function only of the starting and ending points; in particular, if you end at the starting point then the work done is zero. But there are plenty of conservative force fields that still don't satisfy the condition.

The condition is satisfied for all three variables in a uniform force field, in a paraboloid "bowl" force field (radial force everywhere with $F \sim r^2$), and in some other force fields that you might construct with $\mathbf F = f_x(x)\mathbf{\hat x} + f_y(y)\mathbf{\hat y} + f_z(z)\mathbf{\hat z},$ but I can't think of any examples right now (other than the ones already given) that still satisfy the condition for an arbitrary choice of axes.