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\[ \int_V \nabla \cdot {\bf f} \, dV = \int_S {\bf f} \cdot {\bf n} \, dS \]

where the LHS is a volume integral over the volume, \(V\), and the RHS is a surface integral over the surface enclosing the volume. The surface has outward-pointing unit normal, \({\bf n}\). The vector field, \({\bf f}\), can be any vector field at all. Do not assume that it is limited to forces due to the use of the letter \({\bf f}\) in the above equation.

\[ \int_V f_{i,i} \, dV = \int_S f_i n_i \, dS \]

\[ \int_V \, {\partial f_x \over \partial x} + {\partial f_y \over \partial y} + {\partial f_z \over \partial z} \, dV = \int_S f_x n_x + f_y n_y + f_z n_z \, dS \]

But in 1-D, there are no \(y\) or \(z\) components, so we can neglect them. And the volume integral becomes a simple integral over \(x\), so \(dV\) becomes \(dx\).

On the RHS, the surface integral becomes the left and right boundaries on the x-axis, and \(n_x\) equals -1 on the left boundary and +1 on the right. All this reduces the above equation to

\[ \int_{x_1}^{x_2} \, {\partial f_x \over \partial x} \, dx = f(x_2) - f(x_1) \]

And that's it! To show that this works, let \(f(x) = x^2\), then \({\partial f_x \over \partial x} = 2x\), and we get

\[ \int_{x_1}^{x_2} \, 2x \, dx = (x_2)^2 - (x_1)^2 \]

which is clearly the correct 1-D result.

\[ Q = \int_S {\bf v} \cdot {\bf n} \, dS \]

This is easily evaluated because the velocity field is constant, exactly normal to two faces, and exactly parallel to all others. For the left face, \({\bf n} = -1{\bf j}\) and

\[ \int_{Face1} {\bf v} \cdot {\bf n} \, dS = 5 * (-4) = -20 \;\text{m}^3\!\!/\text{s} \]

For the right face, \({\bf n} = 1{\bf j}\) and

\[ \int_{Face2} {\bf v} \cdot {\bf n} \, dS = 5 * (+4) = +20 \;\text{m}^3\!\!/\text{s} \]

So the total integral is

\[ Q = \int_S {\bf v} \cdot {\bf n} \, dS = -20 + 20 = 0 \]

That was easy. But it would be easier still to evaluate

\[ Q = \int_V \nabla \cdot {\bf v} \, dV \]

because since \(\nabla \cdot {\bf v} = 0\), then the integral over any volume of the quantity zero, is zero. And this is exactly equal to the surface integral as it must be.

\[ Q = \int_S {\bf v} \cdot {\bf n} \, dS \]

While this is still possible to solve, it is in fact much easier to apply the divergence theorem and instead evaluate the divergence of the velocity field over the volume.

\[ \nabla \cdot {\bf v} \;\; = \;\; {\partial (5x) \over \partial x} + {\partial (10xz) \over \partial y} + {\partial (-2z) \over \partial z} \;\; = \;\; 3 \]

and the integral of 3 over a volume of 8m

\[ Q \;\; = \;\; \int_V \nabla \cdot {\bf v} \, dV \;\; = \;\; 3 * 8 \;\; = \;\; 24 \]

and this means that the surface integral above must also equal 24. If this were a conservation of mass problem, then the net outflow of material must mean that something very curious is happening to the density!

\[ \int_V \nabla f({\bf x}) \, dV = \int_S f({\bf x}) {\bf n} \, dS \]

where \(f({\bf x})\) is a scalar function of the vector \({\bf x}\). This equation is in fact three separate, independent ones because it is a vector. This is seen more clearly when written in tensor form.

\[ \int_V f,_i \, dV = \int_S f \; n_i \, dS \]

Writing each equation out explicitly gives

\[ \int_V {\partial f({\bf x}) \over \partial x} \, dV = \int_S f({\bf x}) n_x \, dS \qquad \qquad \int_V {\partial f({\bf x}) \over \partial y} \, dV = \int_S f({\bf x}) n_y \, dS \qquad \qquad \int_V {\partial f({\bf x}) \over \partial z} \, dV = \int_S f({\bf x}) n_z \, dS \]

Each equation is separate and can be used independently, in isolation of the others. In fact, the first equation arises in the derivation of the J-Integral. In that case, \(f({\bf x})\) is actually the strain energy density, \(w({\bf x})\), in the vicinity of the crack tip. Finally, note how closely each equation resembles the 1-D case discussed above. Nevertheless, it is a slightly different variation because in this case, the volume is a 3-D object, not 1-D.

A second alternate form involves the application of the divergence theorem to 2nd rank tensors, such as the stress tensor, \(\boldsymbol{\sigma}\).

\[ \int_V \nabla \cdot \boldsymbol{\sigma} \, dV = \int_S \boldsymbol{\sigma } \cdot {\bf n} \, dS \]

This identity often arises because stress-times-area is force. It can be written in tensor notation as

\[ \int_V \sigma_{ij,j} \, dV = \int_S \sigma_{ij} \, n_j \, dS \]

\[ \int_V f,_i \, dV = \int_S f \; n_i \, dS \]

\[ \int_V f_{i,i} \, dV = \int_S f_i n_i \, dS \]

\[ \int_V \sigma_{ij,j} \, dV = \int_S \sigma_{ij} \, n_j \, dS \]

The equations are written for a scalar function, \(f\), and then a vector function, \(f_i\), and finally a tensor function, \(\sigma_{ij}\). In each case, the "\(,i\)" in the volume integral becomes \(n_i\) in the surface integral (except it's a \(j\) in the last example). Tensor notation makes the various forms of the divergence theorem very easy to remember.

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Bob McGinty

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