It's not a volume integral as suggested in the previous posting but a line integral in a pretty awful notation. The correct way is to say we have a curve C, defined, e.g., parametrically as a function \vec{r}:[0,1] \rightarrow \mathbb{R}^3. Then the line integral of a vector field along the curve C is defined as a simple integral over the parameter:
\int_{C} \mathrm{d} \vec{r} \cdot \vec{V}(\vec{r}):=\int_0^1 \mathrm{d} t \; \frac{\mathrm{d} \vec{r}(t)}{\mathrm{d} t} \cdot V[\vec{r}(t)].
Now, sometimes the vector field is "conservative", i.e., the integral doesn't depend on the curve (perhaps restricted to cuves within a certain region of space) but only on the boundary points of the curve. Then you may write the integral in the sloppy way as in the scan of the book.
The specific form of the parametrization chosen is to first integrate from (x_a,y_a,z_a) along a straight line parallel to the x-axis to x_b,y_a,z_a and so on.
This is however, unnecessarily complicated. Since the force field is homogeneous in the example, i.e., \vec{F}(\vec{r})=\text{const} you can integrate along any path connecting the two points, because the line integral is path independent anywhere, because it's analytic everywhere and \vec{\nabla} \times \vec{F}=0 everywhere. Then according to the Poincare lemma, the line integral is path independent. So we can chose just a straight line connecting the points \vec{a} and [/itex]\vec{b}[/itex],
\vec{r}(t)=\vec{a} + t (\vec{b}-\vec{a}) ,\quad t \in [0,1].
This gives
W_{ba}=\int_0^1 \mathrm{d} t (\vec{b}-\vec{a}) \cdot \vec{F}=(\vec{b}-\vec{a}) \cdot \vec{F},
because \vec{F}=\text{const}.
