Surface integrals

Surface integrals of scalar fields

Consider a surface \(S\) in three-dimensional space. The surface integral of its underlying function \(f(x,y,z)\) is defined as \[\begin{equation} \require{physics} \iint_S f(x,y,z) \dd{S} % \thinspace , \end{equation}\] where the integration goes over infinitesimal segments \(\dd{S}\) of the surface. This rather impractical integral can be transformed in a more useful one by parametrizing the surface \[\begin{equation} \mathbf{r}(u,v) % = x(u,v)\mathbf{i} + y(u,v)\mathbf{j} + z(u,v)\mathbf{k} % \thinspace . \end{equation}\] The surface integral can now be written as a double integral over the projected surface \(D\) on an arbitrary plane: \[\begin{equation} \require{physics} \iint_S f(x,y,z) \dd{S} % = \iint_D f(\mathbf{r}(u,v)) % |\mathbf{u}\times\mathbf{v}| % \dd{A} % \thinspace . \end{equation}\]

For the projection on an \(xy\)-plane, the surface is parametrized by \(z=g(x,y)\) as \[\begin{equation} \mathbf{r}(u,v) % = x(u,v)\mathbf{i} + y(u,v)\mathbf{j} + g(x,y)\mathbf{k} % \thinspace . \end{equation}\] The surface integral can now conveniently be written as \[\begin{equation} \require{physics} \iint_S f(x,y,z) \dd{S} % = \iint_D f(x,y,g(x,y)) % \sqrt{ \qty( \pdv{g}{x} )^2 % + \qty( % \pdv{g}{y} % )^2 + 1 % } % \dd{A} % \thinspace . \end{equation}\]

Surface integrals of vector fields

Given a vector field \(\mathbf{F}\) with unit normal vector \(\mathbf{\hat{n}}\), then the surface integral of \(\mathbf{F}\) over surface \(S\) is given by \[\begin{equation} \require{physics} \iint_S \mathbf{F} \vdot \dd{\mathbf{S}} % = \iint_S \mathbf{F} \vdot \mathbf{\hat{n}} \dd{S} % \thinspace . \end{equation}\] This is sometimes called the flux of \(\mathbf{F}\) across \(S\). This equation can easily be interpreted using a simple example. If water is flowing perpendicular to the surface, a lot of water will flow through the surface and the flux will be large. On the other hand, if water is flowing parallel to the surface, there will not be any flux since the water does not flow through the surface. To calculate the total amount of water flowing through the surface, we must take the component of \(\mathbf{F}\) that is perpendicular to the surface. The surface integral can now be evaluated similar to the technique used with scalar fields where the surface is given by \(z=g(x,y)\). Assume that the vector field is given by \[\begin{equation} \mathbf{F} % = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} % \thinspace . \end{equation}\] Under these assumptions, the flux is \[\begin{align} \require{physics} \iint_S \mathbf{F} \vdot d\mathbf{S} % &= \iint_S \mathbf{F} \vdot \mathbf{\hat{n}} \dd{S} % \\ &= \iint_D \mathbf{F} % \vdot \qty( % \frac{ % \mathbf{u} \times \mathbf{v} % }{|\mathbf{u}\times\mathbf{v}|} % ) % |\mathbf{u} \times \mathbf{v}| % \dd{A} % \\ &= \iint_D \qty( % - P\pdv{g}{x} - Q\pdv{g}{y} + R % ) % \dd{A} % \thinspace . \end{align}\] Here, we have, analogous to the scalar counterpart, projected the surface \(S\) onto the \(xy\)-plane and made use of the analytical form of the unit normal vector.