Surface Integrals (Real Scalar Fields)#

TODO: Make more rigorous

Definition: Surface Integral

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^3 \to \mathbb{R}\) be a real scalar field and let \(\phi: \mathcal{D}_{\phi} \subseteq \mathbb{R}^2 \to \mathbb{R}^3\) be a real parametric surface which is totally differentiable, except possibly on a null set of the Lebesgue measure..

The surface integral of \(f\) over \(\phi\) is the integral of the product between \(f \circ \phi\) and magnitude of the normal vector of \(\phi\) over \(\mathcal{D}_{\phi}\) (if it exists):

\[\iint_{\mathcal{D}_{\phi}} f(\phi(u,v)) ||\partial_u \phi(u,v) \times \partial_v \phi(u,v)||\,\mathrm{d}\mathcal{D}_{\phi}\]

Notation

The surface integral of \(f\) over \(\phi\) is typically denoted as follows:

\[\iint_{\phi} f \, \mathrm{d}\phi\]

We also denote \(\mathrm{d}\phi\) in various other ways such as \(\mathrm{d}A\), \(\mathrm{d}S\), etc.

Example

Let \(f: \mathbb{R}^3 \to \mathbb{R}\) be the real scalar field defined as follows:

\[f(x,y,z) = z^2\]

We want to find the surface integral

\[\iint_{\phi} f \,\mathrm{d}\phi\]

of \(f\) over the lateral surface of a cylinder with radius \(R\) and height \(h\) which is parameterized by the following parametric surface:

\[\phi: [0, 2\uppi] \times [0, h] \to \mathbb{R}^3 \qquad \phi(u,v) = \begin{bmatrix}R \cos u \\ R \sin u \\ v\end{bmatrix}\]

For the partial derivatives of \(\phi\), we have:

\[\partial_u \phi(u,v) = \begin{bmatrix} - R \sin u \\ R \cos u \\ 0\end{bmatrix} \qquad \partial_v \phi(u,v) = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\]

The normal vector of \(\phi\) is thus given by

\[\partial_u \phi(u,v) \times \partial_v \phi(u,v) = \begin{bmatrix} - R \sin u \\ R \cos u \\ 0\end{bmatrix} \times \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix}R \cos u \\ R \sin u \\ 0\end{bmatrix}\]

and its magnitude is the following:

\[||\partial_u \phi(u,v) \times \partial_v \phi(u,v)|| = R\]

For the surface integral, we have:

\[\begin{aligned}\iint_{\phi} f \,\mathrm{d}\phi & = \iint_{\mathcal{D}_{\phi}} f(\phi(u,v)) ||\partial_u \phi(u,v) \times \partial_v \phi(u,v)||\,\mathrm{d}\mathcal{D}_{\phi} \\ & = \iint_{[0, 2\uppi] \times [0, h]} R^2 \cos^2 (u) R \, \mathrm{d}\mathcal{D}\end{aligned}\]

Since \(R^2 \cos^2 (u) R\) is continuous, we can evaluate it via an iterated integral:

\[\begin{aligned} \iint_{[0, 2\uppi] \times [0, h]} R^2 \cos^2 (u) R \, \mathrm{d}\mathcal{D} & = \int_0^h \left(\int_0^{2\uppi} R^2 \cos^2 (u) R \,\mathrm{d}u \right) \, \mathrm{d}v \\ & = R^3 \int_0^h \left(\int_0^{2\uppi} \cos^2 u \,\mathrm{d}u \right) \, \mathrm{d}v \\ & = R^3 \int_0^h \left(\int_0^{2\uppi} \frac{1 + \cos(2u)}{2} \,\mathrm{d}u \right) \, \mathrm{d}v \\ & = R^3 \int_0^h \left.\left( \frac{u}{2} + \frac{\sin(2u)}{4} \right)\right\vert_0^{2\uppi} \, \mathrm{d}v \\ & = R^3 \int_0^h \left( \left( \frac{2\uppi}{2} + 0 \right) - (0 + 0) \right) \, \mathrm{d}v \\ & = R^3 \int_0^h \uppi \, \mathrm{d}v \\ & = \uppi R^3 v \vert_0^h \\ & = \uppi R^3 h \end{aligned}\]