4.11 Surface Integrals of Vector Fields
Definition 4.11.1. If \(\mathbf F\) is a continuous vector field defined on an oriented surface \(S\) with unit normal vector \(\mathbf n\), then the surface integral of \(\mathbf F\) over \(S\) is \[\iint \limits _S \mathbf F\cdot d\mathbf S = \iint \limits _S \mathbf F\cdot \mathbf n\, dS .\] This integral is also called the flux of \(\mathbf F\) across \(S\).
If \(\mathbf F\) is the velocity field of a fluid, the flux is the volume of fluid crossing \(S\) per unit time, in the direction of \(\mathbf n\). If \(S\) is given by a vector function \(\mathbf r(u,v)\) with the orientation \(\mathbf n = \frac {\mathbf r_u\times \mathbf r_v}{|\mathbf r_u\times \mathbf r_v|}\), then \[\iint \limits _S \mathbf F\cdot d\mathbf S = \iint \limits _D \mathbf F(\mathbf r(u,v))\cdot \frac {\mathbf r_u \times \mathbf r_v}{|\mathbf r_u \times \mathbf r_v|}\, |\mathbf r_u \times \mathbf r_v|\,dA = \iint \limits _D \mathbf F(\mathbf r(u,v))\cdot (\mathbf r_u \times \mathbf r_v) \,dA .\]
Example 4.11.2. Find the flux of the vector field \(\mathbf F(x,y,z) = z\,\mathbf i + y\,\mathbf j + x\, \mathbf k\) across the unit sphere \(x^2 + y^2 + z^2 = 1\), oriented outward.
Solution. With the spherical parametrisation, \(\mathbf F(\mathbf r(\phi ,\theta )) = \cos \phi \, \mathbf i + \sin \phi \sin \theta \,\mathbf j + \sin \phi \cos \theta \,\mathbf k\) and \(\mathbf r_\phi \times \mathbf r_\theta = \sin ^2 \phi \cos \theta \, \mathbf i + \sin ^2\phi \sin \theta \, \mathbf j + \sin \phi \cos \phi \, \mathbf k\), which points outward. Their dot product is \[\cos \phi \sin ^2 \phi \cos \theta + \sin ^3\phi \sin ^2\theta + \sin ^2\phi \cos \phi \cos \theta = 2\sin ^2\phi \cos \phi \cos \theta + \sin ^3\phi \sin ^2\theta ,\] so \[\iint \limits _S \mathbf F\cdot d\mathbf S = \int ^{2\pi }_0\int ^{\pi }_0 \left (2\sin ^2\phi \cos \phi \cos \theta + \sin ^3\phi \sin ^2\theta \right )d\phi \, d\theta = 0 + \frac 43\cdot \pi = \frac {4\pi }{3},\] the first term vanishing because \(\int _0^{2\pi }\cos \theta \,d\theta = 0\).
For a surface \(S\) given as \(z = g(x,y)\) over \(D\) and oriented upward, with \(\mathbf F = P\,\mathbf i + Q\, \mathbf j + R\, \mathbf k\), \[\iint \limits _S \mathbf F\cdot d\mathbf S = \iint \limits _D \left ( -P\frac {\partial g}{\partial x}- Q \frac {\partial g}{\partial y} + R\right )dA ,\] since then \(\mathbf r_x\times \mathbf r_y = \langle -g_x, -g_y, 1\rangle \).
Example 4.11.3. Find the flux of \(\mathbf F = y\,\mathbf i + x\,\mathbf j + z\,\mathbf k\) upward through the part of the paraboloid \(z = 1 - x^2 - y^2\) above the \(xy\)-plane.
Solution. With \(g = 1 - x^2 - y^2\), \(g_x = -2x\), \(g_y = -2y\), so the integrand is \(-y(-2x) - x(-2y) + (1 - x^2 - y^2) = 4xy + 1 - x^2 - y^2\). Over the unit disc the term \(4xy\) integrates to zero by symmetry, and \[\iint \limits _S\mathbf F\cdot d\mathbf S = \int _0^{2\pi }\int _0^1(1 - r^2)\,r\, dr\,d\theta = \frac {\pi }{2} .\]
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