To understand the various operations in vector calculus, it's good to have the intuitive example of fluid dynamics (i.e., the theory of the flow of gases or liquids) at hand. It's the most simple example for a classical field theory which makes sense in both the non-relativistic and the relativistic realm. Let's concentrate on the non-relativistic case for now.
To characterize the fluid, we introduce some of the most important physical quantities related with it. We usually describe the fluid in terms of Euler coordinates. Here we have in mind an observer, who looks at the momentary state of the fluid at a given position \vec{x} as a function of time. It is in important to keep in mind that in this picture at any instant of time a different portion of fluid molecules is located at this point.
To describe the motion of the fluid, we introduce the velocity field, i.e., the momentary velocity of the fluid cell at any position, \vec{v}(t,\vec{x}). Another important quantity is the density of the fluid at any position, \rho(t,\vec{x}). This gives the mass per volume of a little fluid element at this position as a function of time.
The first law we want to describe is the conservation of mass (which is a necessary conclusion from the space-time properties of Newtonian mechanics, i.e., a very fundamental law). It tells us that a given amount of fluid never changes its total mass. To see, how to formulate the corresponding law we put a fixed time-independent volume V into the fluid. Then the mass-conservation law tells us that the mass of the matter contained in it can only change by the flow of fluid particles into or out of this volume through its boundary \partial V. For this boundary we also have to specify the orientation relative to the volume enclosed by it. The standard choice is to direct each area-normal vector \mathrm{d}^2 \vec{A} out of the volume V.
Then through such a surface element in an infinitesimal time \mathrm{d} t there flows the amount of mass that is contained in the volume element \mathrm{d} t \mathrm{d} \vec{F} \cdot \vec{v}(t,\vec{x}). Fluid flowing out has to be counted negative, so we have
\mathrm{d} m_V=\mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} \int_V \mathrm{d}^3 \vec{x} \rho(t,\vec{x})=-\mathrm{d} t \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \rho(t,\vec{x}) \vec{v}(t,\vec{x}).
Using Gauss's integral theorem we can write
\int_V \mathrm{d}^3 \vec{x} \partial_t \rho(t,\vec{x})=-\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \rho(t,\vec{x}) \vec{v}(t,\vec{x})=-\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot [\rho(t,\vec{x}) \vec{v}(t,\vec{x})].
To also understand the curl operation we investigate, how the relative vectors of two close fluid cells within the fluid change with time. In the infinitesimal time increment \mathrm{d} t each point changes its position by
\vec{\Delta}(t,\vec{x})=\vec{v}(t,\vec{x}) \mathrm{d} t.
The relative deformation for two infinitesimally close points \vec{x} and \vec{x}+\mathrm{d} x thus is
\mathrm{d} \vec{x} \cdot \vec{\nabla} \vec{\Delta}=\vec{e}_k \mathrm{d} t \mathrm{d} x_j \partial_j v_k.
Now we can write
\partial_j v_k=\frac{1}{2} (\partial_j v_k-\partial_k v_j)+\frac{1}{2} (\partial_j v_k+\partial_k v_j)=\epsilon_{jkl} \omega_{j} + \epsilon_{jk}.
Now let's investigate the change of the distance of the two adjacent points:
\mathrm{d} l(t)=|\mathrm{d} \vec{x}|, \quad \mathrm{d} l(t+\mathrm{d} t)=\sqrt{(\mathrm{d} \vec{x}+\Delta)^2}=\mathrm{d} l(t)+\mathrm{d} \vec{x} \cdot \Delta + \mathcal{O}(\mathrm{d}t^2)=\mathrm{d} l(t) + \mathrm{d} x_k \mathrm{d} x_j \partial_j v_k \mathrm{d} t=\mathrm{d} l(t)+\epsilon_{jk} \mathrm{d}x_j \mathrm{d} x_k \mathrm{d} t.
The part of the motion due to the antisymmetric part thus corresponds to an infinitesimal rotation, because it doesn't give rise to a length change.
\vec{\Delta}_{\text{rot}}=\mathrm{d} t \vec{\omega} \times \mathrm{d} \vec{x},
of the fluid cell as a whole and thus is not part of an intrinsic deformation.
In addition usually the fluid cell can be moving as a whole in a parallel motion. This "rigid translation" is canceled completely in considering the relative vector
The intrinsic deformation of the fluid element is totally due to the symmetric part \epsilon_{jk} of the derivatives of the velocity field. This symmetric strain tensor can be decomposed further in a homogeneous part and a traceless part,
\epsilon_{jk}=\frac{1}{3} \epsilon_{ll} \delta_{jk} + \eta_{jk}.
To analyze the physical meaning of this decomposition, we evaluate the change of the volume of the infinitesimal fluid element.
To that end we remember that we can always choose a Cartesian coordinate system such that the symmetric deformation tensor \epsilon_{jk} becomes diagonal. An infinitesimal box of fluid parallel to the corresponding principal coordinate axes changes its volume according to
\mathrm{d} V(t+\mathrm{d} t)=(\mathrm{d} x_1+\Delta_1)(\mathrm{d} x_2+\Delta_2) (\mathrm{d} x_3+\Delta_3)=\mathrm{d} V(t) [1+(\partial_1 v_1 + \partial_2 v_2 + \partial_3 v_3) \mathrm{d} t]+\mathcal{O}(\mathrm{d} t^2)=\mathrm{d} V(t) [1+\mathrm{d} t \vec{\nabla} \cdot \vec{v}]+\mathcal{O}(\mathrm{d} t^2)=\mathrm{d} V(t) [1+\mathrm{d} t \mathrm{tr} \epsilon]+\mathcal{O}(\mathrm{d} t^2).
Now the trace is an invariant under coordinate transformations and thus can be evaluated in any coordinate system. Thus the trace of the strain tensor describes a change of the volume.
The remaining deformation of the fluid element doesn't change the volume and is thus to be interpreted as a shear.
This completely visualizes the various covariant first-order derivatives of the velocity field.
