4.14 The divergence Theorem
The divergence theorem is the version of Green’s theorem one dimension up: the flux of a field out of a closed surface equals the integral of its divergence over the solid inside.
Theorem 4.14.1 (Divergence theorem). Let \(E\) be a simple solid region and let \(S\) be the boundary surface of \(E\), with positive (outward) orientation. Let \(\mathbf F\) be a vector field whose component functions have continuous partial derivatives on an open region that contains \(E\). Then \[\iint \limits _S \mathbf F\cdot d\mathbf S = \iiint \limits _E \dive \mathbf F\, dV .\]
The theorem justifies the description of the divergence as outflow per unit volume: applied to a small ball \(B_\varepsilon \) about a point \(P\), \(\iint _{\partial B_\varepsilon }\mathbf F\cdot d\mathbf S\approx \dive \mathbf F(P)\cdot \text {vol}(B_\varepsilon )\).
Example 4.14.2. Find the flux of the vector field \(\mathbf F = z\,\mathbf i + y\,\mathbf j + x\,\mathbf k\) over the unit sphere.
Solution. \(\dive \mathbf F = 0 + 1 + 0 = 1\). Let \(B\) be the unit ball. By the divergence theorem, \[\iint \limits _S \mathbf F\cdot d\mathbf S = \iiint \limits _B \dive \mathbf F\,dV = \iiint \limits _B dV = \text {vol}(B) = \frac {4\pi }{3},\] in agreement with the direct calculation of Example 4.11.2.
Example 4.14.3. Use the divergence theorem to find the flux of Example 4.12.1, and the outward flux of \(\mathbf F = x^3\,\mathbf i + y^3\,\mathbf j + z^3\,\mathbf k\) through the sphere of radius \(a\).
Solution. For the cube, \(\dive \mathbf F = 1 + 2 + 3 = 6\), and the volume is \(1\), so the flux is \(6\), as found face by face. For the second field, \(\dive \mathbf F = 3(x^2 + y^2 + z^2) = 3\rho ^2\), so \[\iint \limits _S\mathbf F\cdot d\mathbf S = \int _0^{\pi }\int _0^{2\pi }\int _0^a 3\rho ^2\,\rho ^2\sin \phi \,d\rho \,d\theta \,d\phi = 3\cdot \frac {a^5}{5}\cdot 2\pi \cdot 2 = \frac {12\pi a^5}{5} .\]
Example 4.14.4. Evaluate \(\iint _S \mathbf F\cdot d\mathbf S\), where \(\mathbf F = xy\,\mathbf i + \left (y^2 + e^{xz^2}\right )\mathbf j + \sin (xy)\,\mathbf k\) and \(S\) is the surface of the region \(E\) bounded by the parabolic cylinder \(z = 1 - x^2\) and the planes \(z = 0\), \(y = 0\) and \(y + z = 2\).
Solution. Computing four surface integrals directly would be very hard, but \(\dive \mathbf F = y + 2y + 0 = 3y\). The region is \(-1\leq x\leq 1\), \(0\leq z\leq 1 - x^2\), \(0\leq y\leq 2 - z\), so \begin {align*} \iint \limits _S\mathbf F\cdot d\mathbf S & = \int _{-1}^1\int _0^{1 - x^2}\int _0^{2 - z}3y\,dy\,dz\,dx = \frac 32\int _{-1}^1\int _0^{1 - x^2}(2 - z)^2\,dz\,dx\\ & = \frac 12\int _{-1}^1\left [8 - (x^2 + 1)^3\right ]dx = \frac {184}{35} . \end {align*}
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