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Advanced Engineering Mathematics Lecture Notes: West Virginia Wesleyan College

Jesse Oldroyd

Section 4.5 The Divergence Theorem and Its Applications

In Section 4.3, we discussed Green's Theorem for flux integrals. This result allows us to relate the flux integral of a vector field \(\FF\) along the boundary \(\partial R\) of a simple, closed and bounded region \(R\) in the plane with the double integral of the divergence. We can do something similar in \(\RR^3\) by replacing the flux integral with the surface (flux) integral. We will see later that this can be viewed as a kind of “integration by parts” formula in higher dimensions.

Subsection The Divergence Theorem

We begin by stating the Divergence Theorem in \(\RR^3\text{.}\) Unless we mention otherwise, all surface normal vectors will be assumed to be outward-pointing relative to the region enclosed by the surface.
The triple integral that appears in the Divergence Theorem can be viewed as the “net outflow” of \(\FF\) over the points of \(T\text{.}\) From this viewpoint, the Divergence Theorem states that this net outflow is exactly equal to the flux of \(\FF\) out of the boundary of \(T\text{.}\)

Example 4.5.2. Flux Out of a Hemisphere.

Compute \(\iint_D \FF\cdot\dd{\SS}\) where
\begin{equation*} \FF = \mqty[x^3 - y^3 \amp y^3 - z^3 \amp z^3 - x^3] \end{equation*}
and \(D\) denotes the upper hemisphere of the sphere of radius \(5\) centered at the origin.
Solution.
Using methods from Section 4.4, we would need to split the piecewise smooth surface \(D\) into two smooth components \(D_1\) and \(D_2\) representing the spherical cap and the base of \(D\) respectively. See the plot produced below. Note that the plot also indicates that our surface integral should be positive.
This is manageable but tedious, so we instead use Theorem 4.5.1 to rewrite the surface integral as a single triple integral,
\begin{equation*} \iint_D\FF\cdot\dd{\SS} = \iiint_T 3(x^2 + y^2 + z^2)\dd{V}\text{,} \end{equation*}
where \(T\) denotes the region enclosed within the hemisphere. To evaluate this integral we convert to spherical coordinates:
\begin{align*} \iiint_T 3(x^2+y^2+z^2)\dd{V} \amp = 3\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^5\rho^4\sin(\varphi)\dd{\rho}\dd{\varphi}\dd{\theta} \\ \amp = 3750\pi\text{.} \end{align*}
Our answer is positive, which is consistent with the graph produced above.

Example 4.5.3. Flux Into a Cube.

Let \(\FF = \mqty[xy \amp y^2 z \amp z^3] \text{.}\) Find the flux of \(\FF\) into the unit cube \(T: 0\leq x, y, z\leq 1\text{.}\)
Solution.
If we try to compute this without using the Divergence Theorem then we would need to compute six separate surface integrals, one for each face of the cube. If we make use of the Divergence Theorem, this problem becomes much more tractable. So let \(T\) denote the unit cube and the region it encloses. Then
\begin{align*} -\iint_{\partial T} \FF\cdot\dd{\SS} \amp = -\int_0^1\int_0^1\int_0^1 (y + 2yz + 3z^2)\dd{z}\dd{y}\dd{x}\\ \amp = -2 \text{.} \end{align*}
Therefore \(\FF\) tends to flow out of the cube.

Subsection Applications of the Divergence Theorem

Theorem 4.5.1 leads to several other integral formulas that can be useful in higher dimensions. For our first example, we'll consider what happens if \(\FF\) is conservative. If this is the case, then there exists a potential function \(f\) for \(\FF\text{:}\) \(\del{f} = \FF\text{.}\) Then
\begin{align*} \div\FF \amp = \div\del{f} = \Delta f\\ \FF\cdot\dd{\SS} \amp = \del{f}\cdot\nn\dd{S} \end{align*}
where \(\Delta f = f_{xx} + f_{yy}\) is the Laplacian of \(f\) and \(\del{f}\cdot\nn\) is the normal derivative of \(f\text{,}\) often denoted by \(\pdv{f}{n}\text{.}\)

Example 4.5.4. Green's First Identity.

Let \(T\) satisfy the conditions of Theorem 4.5.1 and let \(f(x, y, z)\) and \(g(x, y, z)\) be twice continuously differentiable functions. Then
\begin{equation*} \iiint_T(f\Delta g + \del{f}\cdot\del{g})\dd{V} = \iint_{\partial T} f\pdv{g}{n}\dd{S}\text{.} \end{equation*}
Solution.
To prove this we'll set \(\FF = f\del{g}\text{.}\) Then
\begin{equation*} \div\FF = \del{f}\cdot\del{g} + f\Delta g \end{equation*}
and
\begin{equation*} \FF\cdot\nn = f(\del{g}\cdot\nn) = f\pdv{g}{n}\text{,} \end{equation*}
and the Divergence Theorem then implies the result.
Green's First Identity is particularly important for proving a result about harmonic functions, which are scalar functions \(f\) such that \(\Delta f = 0\text{.}\) If we set \(f = g\) in the identity where \(f\) is harmonic then we get
\begin{equation*} \iiint_T \norm{\del f}^2\dd{V} = \iint_{\partial T}f\pdv{f}{n}\dd{S}\text{.} \end{equation*}
If we also know that \(f = 0\) everywhere on \(\partial T\text{,}\) then this forces \(\del{f} = \mathbf{0}\) and furthermore \(f = 0\text{.}\) This gives us the following result.
Set \(h = f - g\) and note that \(h\) is harmonic on \(T\) and that \(h = 0\) on \(\partial T\text{.}\) Therefore \(h = 0\) inside of \(T\) as well, which implies that \(f = g\) inside of \(T\text{.}\)
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