4.11 Surface Integrals of Vector Fields
If \(F\) is a continuous vector field defined on an oriented surface \(S\) with normal vector \(\widehat {n}\), then the surface integral of \(F\) over \(S\) is \[ \iint \limits _S F\cdot dS = \iint \limits _S F\cdot \widehat {n}\hspace {0.1cm} dS\]
This integral is also called the flux of \(F\) across \(S\).
If \(S\) is given by a vector function \(\overline {r}(u,v)\), then \(\displaystyle {\widehat {n}=\frac {\overline {r}_u \times \overline {r}_v}{\big |\overline {r}_u \times \overline {r}_v\big |}}\).
Hence, \begin {align*} \iint \limits _S F\cdot ds & = \iint \limits _S F \cdot \frac {\overline {r}_u \times \overline {r}_v}{\big |\overline {r}_u \times \overline {r}_v\big |}\hspace {0.1cm} dS\\\\ & = \iint \limits _D F\big [\overline {r}(u,v)\big ]\cdot \frac {\overline {r}_u \times \overline {r}_v}{\big |\overline {r}_u \times \overline {r}_v\big |}\hspace {0.1cm} \big |\overline {r}_u \times \overline {r}_v\big |\hspace {0.1cm}dA\\\\ \therefore \hspace {0.5cm}\iint \limits _S F\cdot ds & = \iint \limits _D F\big [\overline {r}(u,v)\big ]\cdot \big (\overline {r}_u \times \overline {r}_v\big ) \hspace {0.1cm}dA\\\\ \end {align*}
Find the flux of the vector field \(F(x,y,z) = z\,\textbf {i} + y\,\textbf {j} + x\, \textbf {k}\) across the unit sphere \(x^2 + y^2 + z^2 = 1\).
Solution.
\(\overline {r}(\phi , \theta ) = \sin \phi \cos \theta \, \textbf {i} + \sin \phi \sin \theta \, \textbf {j} + \cos \phi \, \textbf {k}\)
\[F\big [\overline {r}(\phi ,\theta )\big ] = \cos \phi \, \textbf {i} + \sin \phi \sin \theta \,\textbf {j} + \sin \phi \cos \theta \,\textbf {k}\]
\[\overline {r}_{\phi } \times \overline {r}_{\theta } = \sin ^2 \phi \cos \theta \, \textbf {i} + \sin ^2\phi \sin \theta \, \textbf {j} + \sin \phi \cos \phi \, \textbf {k}\]
\[F\big [\overline {r}(\phi ,\theta )\big ] \cdot \big (\overline {r}_{\phi } \times \overline {r}_{\theta }\big )= \cos \phi \sin ^2 \phi \cos \theta + \sin ^3\phi \sin ^2\theta + \sin ^2\phi \cos \theta \cos \phi \]
\begin {align*} \iint \limits _F F\cdot dS & = \iint \limits _D F\cdot \big (\overline {r}_{\phi } \times \overline {r}_{\theta }\big ) dA\\\\ & = \int ^{2\pi }_0\int ^1_0\big [2\sin ^2\phi \cos \theta + \sin ^3\phi \sin ^2\theta \big ]\hspace {0.1cm}d\phi \hspace {0.1cm}d\theta \\\\ & = \frac {4\pi }{3}\\\\ \end {align*}
In the case of a surface \(S\) given as \(z = g(x,y)\), \[\iint \limits _S F\cdot dS = \iint \limits _D \Bigg ( -P\dfrac {\partial g}{\partial x}- Q \frac {\partial g}{\partial y} + R\Bigg )dA\] where \(F = P\,\textbf {i} + Q\, \textbf {j} + R\, \textbf {k}\).
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