I am attempting to generalize the equation for the total work done by multiple, constant forces on a system that can be modeled as a single particle (that is, a system that moves so that all the parts of the system undergo identical displacements). My primary source is Chapter 6 of Tipler and Mosca's Physics For Scientists and Engineers, 6th Edition. If $n$ constant forces do work on a system of particles constrained to move in one dimension, then the total work is found by computing the work done by each force and adding each individual work together: $$W_\text{total}=W_1+W_2+\cdots+W_n=F_{1x}\Delta x_1+F_{2x}\Delta x_2+\cdots+F_{nx}\Delta x_n=\sum_{i=1}^n F_{ix}\Delta x_i \tag{1}\label{eq1}$$ Now, if we allow the forces to be variable and for the particles of the system to move in all three spatial dimensions, it should still be true that the total work is equal to the sum of the individual work done by each force acting on a particle in the system, except that now the appropriate formula is $$W_\text{total}=\int_{C_1}\mathbf{F}_1\cdot d\mathbf{r}+\int_{C_2}\mathbf{F}_2\cdot d\mathbf{r}+\cdots+\int_{C_n}\mathbf{F}_n\cdot d\mathbf{r}=\sum_{i=1}^n\int_{C_i}\mathbf{F}_i\cdot d\mathbf{r} \tag{2}\label{eq2}$$ where $\mathbf{r}=x\hat{\mathbf{i}}+y\hat{\mathbf{j}}+z\hat{\mathbf{k}}$ is the displacement of the particle and $C_i$ is the smooth curve traversed by the particle upon which $\mathbf{F}_i=F_{ix}\hat{\mathbf{i}}+F_{iy}\hat{\mathbf{j}}+F_{iz}\hat{\mathbf{k}}$ acts. Assuming that this is all correct, is it possible to simplify this equation in the case that the particles that comprise the system undergo identical displacements so that the system can be modeled as a single particle? Given that the work done by constant forces on a system of particles that can be modeled as a single particle and is constrained to move in one dimension is $$W_\text{total}=F_{1x}\Delta x+F_{2x}\Delta x+\cdots+F_{nx}\Delta x=\left(\sum_{i=1}^n F_{ix}\right)\Delta x=F_{\text{net}\,x}\Delta x \tag{3}\label{eq3}$$
my first thought was that for the work done by variable forces on system of particles that can be modeled as a single particle and is allowed to move in three spatial dimensions, the fact that the displacements of the particles are identical means that the $C_i$ are identical so that $\eqref{eq2}$ becomes $$W_\text{total}=\int_{C}\mathbf{F}_1\cdot d\mathbf{r}+\int_{C}\mathbf{F}_2\cdot d\mathbf{r}+\cdots+\int_{C}\mathbf{F}_n\cdot d\mathbf{r}=\int_C(\mathbf{F}_1+\mathbf{F}_2+\cdots+\mathbf{F}_n)\cdot d\mathbf{r}$$ $$=\int_{C}\left(\sum_{i=1}^n\mathbf{F}_i\right)\cdot d\mathbf{r}=\int_C \mathbf{F}_\text{net}\cdot d\mathbf{r} \tag{4}\label{eq4}$$ where $C$ is the smooth curve traversed by any one of the particles that comprise the system. But the more I think about it, I feel that this formulation may be incorrect. While the particles of the system undergo identical displacements, the initial and final positions of the particles of the system must be different. (Otherwise, all the particles would occupy a singularity in space and would therefore be the same particle.) Because the forces acting upon the system may vary in three-dimensional space, making the assumption that the $C_i$ are identical may result in the work done by one or more of the variable forces being incorrectly evaluated as the particles of the system move through different regions of space during their journey (even though their overall displacements are identical). Is this line of reasoning correct? If so, is there another way to approach the generalization of $\eqref{eq3}$/the simplification of $\eqref{eq2}$ to a system of particles that can be modeled as a single particle being acted on by multiple variable forces (in three dimensions)? If so, what would it be? Also, if the forces acting on the system (modeled as a single particle) are all constant, is it possible to express the total work as $$W_\text{total}=\mathbf{F}_1\cdot\mathbf{r}+\mathbf{F}_2\cdot\mathbf{r}+\cdots+\mathbf{F}_n\cdot\mathbf{r}=\left(\mathbf{F}_1+\mathbf{F}_2+\cdots+\mathbf{F}_n\right)\cdot\mathbf{r}=\sum_{i=1}^n\mathbf{F}_i\cdot\mathbf{r}=\mathbf{F}_\text{net}\cdot\mathbf{r} \tag{5}\label{eq5}$$
