Fundamentals of Continuum Mechanics

Problem 9.4

Prove the transport theorem for surface integrals. That is, show that

$\frac{\text{d}}{\text{d}t}{\displaystyle \underset{\mathcal{Q}}{\int }\mathbf{a}·\mathbf{n}\text{d}a}={\displaystyle \underset{\mathcal{Q}}{\int }\left[{\mathbf{a}}^{\prime }+\text{curl}\left(\mathbf{a}\times \mathbf{v}\right)+\mathbf{v}\left(\text{div}\mathbf{a}\right)\right]}·\mathbf{n}\text{d}a.$

Solution

We begin by using relationship (3.76) between n da and N dA, together with a change of independent variable from x to X, to convert the Eulerian integration to a Lagrangian integration, i.e.,

$\frac{\text{d}}{\text{d}t}{\displaystyle \underset{\mathcal{Q}}{\int }\mathbf{a}·\mathbf{n}\text{d}a}=\frac{\text{d}}{\text{d}t}{\displaystyle \underset{{\mathcal{Q}}_{\text{R}}}{\int }\mathbf{a}·J{\mathbf{F}}^{-\text{T}}\mathbf{N}\text{d}A}.$

We emphasize that the integrand on the left-hand side is a function of x and t, while the integrand on the right-hand side is a function of X and t. Then, it follows that

$\begin{array}{ll}\frac{\text{d}}{\text{d}t}{\displaystyle \underset{\mathcal{Q}}{\int }\mathbf{a}·\mathbf{n}\text{d}a}\hfill & =\frac{\text{d}}{\text{d}t}{\displaystyle \underset{{\mathcal{Q}}_{\text{R}}}{\int }J{\mathbf{F}}^{-1}}\hfill \end{array}$