8.4 Integral Theorems
Gauss' Integral Theorem.
Let \( V \) be a volume with surface A and \( \mathbf{n} \) an outward unit
vector on \( A \). The for any field \( f(x) \):
$$
\begin{equation}
\int_V f_{,k} \; dV = \int_A f \, n_k \; dA
\tag{8.30}
\end{equation}
$$
For a proof see e.g. [2] (C.3).
The Divergence Theorem.
By replacing the function \( f(x) \) in by the vector components \( a_i (x) \)
of a vector field \( \mathbf{a}(x) \) we get:
$$
\begin{align}
\int_V a_{k,k} \; dV &= \int_A a_k \, n_k \; dA \tag{8.31}\\
\int_V \nabla \cdot \mathbf{a}\; dV &= \int_A \mathbf{a \cdot n}\; dA
\tag{8.32}
\end{align}
$$
It is trivial that Eq. (8.31) holds when no summation
involved, by just replacing \( f \) with \( a_k \) in
Eq. (8.30). However, as the integral operator is linear, it will
also hold when Einstein's summation convention is
employed. Consequently, Eq. (8.31) is valid when using the
summation convention too, and then Eq. (8.32) follows as
the vector representation of Eq. (8.31).
The First Mean Value Theorem.
Let \( f(\mathbf{x}) \) and \( g(\mathbf{x}) \) represent continuous field
functions, such that \( f,g : \mathbb{R}^3 \rightarrow \mathbb{R} \).
Further, let \( V \) represent a contiguous volume, i.e., \( V \in
\mathbb{R}^3 \). Then there exists a \( \bar{\mathbf{x}} \in V \) such
that:
$$
\begin{equation}
\int_V f(\mathbf{x}) \, g(\mathbf{x}) \; dV = f(\bar{\mathbf{x}}) \int_V
\, g(\mathbf{x}) \; dV
\tag{8.33}
\end{equation}
$$
In particular, if \( g(\mathbf{x}) =1 \) for all \( \mathbf{x} \in V \)
$$
\begin{equation}
\int_V f(\mathbf{x}) \; dV = f(\bar{\mathbf{x}}) \, V
\tag{8.34}
\end{equation}
$$
Thus, \( f(\bar{\mathbf{x}}) \) is the mean value of \( f(\mathbf{x}) \) in
\( V \).
For proofs see [48] and [2] (C.3).
Leibniz's rule for 1D integrals.
Let
$$
\begin{equation}
F(t) = \int_{a(t)}^{b(t)} f(x,t) \, dx
\tag{8.35}
\end{equation}
$$
then
$$
\begin{equation}
\frac{dF(t)}{dt} = \int_{a(t)}^{b(t)} \partd{f(x,t)}{t} \, dx +
f(x,t) \partd{b(t)}{t} - f(x,t) \partd{a(t)}{t}
\tag{8.36}
\end{equation}
$$
Leibniz's rule for 2D integrals.
Let us consider the integral \( F(t) \) obtained by integration
of the function \( f(x,y,t) \) over the domain in 2D space \( \{x,y\} \in A(t) \)
where \( A(t) \) is a function of \( t \):
$$
\begin{equation}
F(t) = \int_{A(t)} f(x,y,t) \, dA
\tag{8.37}
\end{equation}
$$
then:
$$
\begin{equation}
\frac{dF(t)}{dt} = \int_{A(t)} \partd{f(x,y,t)}{t} \, dA + \left [
f(x,y,t) \frac{dA}{dt} \right ]_{C(t)} =
\int_{A(t)} \partd{f(x,y,t)}{t} \, dA + \oint_{C(t)} f \frac{d{n}}{dt}
\, d{c}
\tag{8.38}
\end{equation}
$$
where \( dA \) at the boundary (or contour) is expressed as \( dA= dn \,
dc \). Hence, \( (n,c) \) is the local coordinate, where \( dc \) is tangential
and \( dn \) is orthogonal to \( C(t) \).
If \( \boldsymbol{v} \) is the velocity on \( C(t) \), the velocity component of
\( \boldsymbol{v} \) in the outward normal direction \( \boldsymbol{n} \) to \( C(t) \) is:
$$
\begin{equation}
{v}_n = \boldsymbol{v} \cdot \boldsymbol{n} = \frac{dn}{dt}
\tag{8.39}
\end{equation}
$$
and thus:
$$
\begin{equation}
\frac{d{n}}{dt} \, d{c} = v_n \, dc = \boldsymbol{v} \cdot \boldsymbol{n} \, dc
\tag{8.40}
\end{equation}
$$
and consequently:
$$
\begin{equation}
\frac{dF(t)}{dt} = \int_{A(t)} \partd{f(x,y,t)}{t} \, dA +
\oint_{C(t)} f \, \boldsymbol{v}\cdot \boldsymbol{n} \, dc
\tag{8.41}
\end{equation}
$$
Leibniz's rule for 3D integrals.
Let us consider the integral \( F(t) \) obtained by integration
of the function \( f(x,y,z,t) \) over the domain in 3D space \( \{x,y,z\} \in V(t) \)
where \( V(t) \) is a function of \( t \):
$$
\begin{equation}
F(t) = \int_{V(t)} f(x,y,z,t) \, dV
\tag{8.42}
\end{equation}
$$
The, volume \( V(t) \) has the surface \( A(t) \).
$$
\begin{equation}
\frac{dF(t)}{dt} = \int_{V(t)} \partd{f(x,y,t)}{t} \, dV + \left [
f(x,y,t) \frac{dV}{dt} \right ]_{A(t)} =
\int_{V(t)} \partd{f(x,y,t)}{t} \, dV + \int_{A(t)} f \frac{d{n}}{dt}
\, d{A}
\tag{8.43}
\end{equation}
$$
where \( dV= dn \, dA \) is the area increment and \( dn \) is the increment
in the direction of the outward normal to \( A \).
Note that:
$$
\begin{equation}
\frac{dV}{dt} = \frac{dn}{dt} \, dA = v_n \, dA = \boldsymbol{v} \cdot
\boldsymbol{n} \, dA
\tag{8.44}
\end{equation}
$$
and consequently:
$$
\begin{equation}
\frac{dF(t)}{dt} = \int_{V(t)} \partd{f(x,y,t)}{t} \, dV +
\int_{A(t)} f \, \boldsymbol{v} \cdot \boldsymbol{n} \, d{A}
\tag{8.45}
\end{equation}
$$