Surface Integrals and Flux

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We now define the integral over a surface by adding one more variable to the line integral.

Parametrized Surfaces

Definition 1 A \(C^1\) map defined on a planar region \(D\)

\(\mathbf{r}\colon D \rightarrow \mathbb{R}^3\), \(\mathbf{r}(u, v) = (x(u,v), y(u,v), z(u,v))\)

is called a parametrized surface.

Fixing \(u\) and varying \(v\) traces a curve on the surface whose tangent is \(\mathbf{r}_v\); likewise, \(\mathbf{r}_u\) is the tangent to another curve on the surface. The plane spanned by these two tangent vectors is the tangent plane, and \(\mathbf{r}_u \times \mathbf{r}_v\), being perpendicular to it, gives the normal direction. Thus the partial-derivative vectors \(\mathbf{r}_u = \partial \mathbf{r}/\partial u\) and \(\mathbf{r}_v = \partial \mathbf{r}/\partial v\) are tangent to the surface, and their cross product

\[\mathbf{N} = \mathbf{r}_u \times \mathbf{r}_v\]

is the normal vector of the surface. We call a parametrized surface regular if \(\mathbf{N}(u,v) \neq \mathbf{0}\) for all \((u,v) \in D\).

Surface Area

Dividing the surface into small rectangles in the parameter domain, each piece is approximated by a small parallelogram on the tangent plane, more precisely by the parallelogram spanned by \(\mathbf{r}_u\Delta u\) and \(\mathbf{r}_v\Delta v\). Since its area is \(\lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\Delta u\Delta v\), summing these and passing to the limit yields the surface area.

Definition 2 The surface area of a regular parametrized surface \(\mathbf{r}\colon D \rightarrow \mathbb{R}^3\) is

\[\iint_D \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert \dd{u}\dd{v}\]

and we write the area element as \(\dd{S} = \lvert \mathbf{r}_u \times \mathbf{r}_v\rvert \dd{u}\dd{v}\).

The parametrizations we shall use often fail to satisfy the regularity condition on all of \(D\). Even if \(\mathbf{N}\) vanishes at finitely many points and curves, or if distinct parameters map to the same point, the exceptional set has area \(0\) and therefore does not contribute to the value of the above double integral. Hence, when we speak of surface area and surface integrals hereafter, we allow parametrizations that are regular and injective except for finitely many points and curves, that is, parametrizations that cover the surface exactly once.

The area element \(\dd{S}\) plays the same role as the Jacobian determinant in multiple integrals, and with this definition we can integrate a scalar quantity distributed over the surface.

Definition 3 The surface integral of a continuous scalar field \(f\) on the image \(S = \mathbf{r}(D)\) of a regular parametrized surface \(\mathbf{r}\colon D \rightarrow \mathbb{R}^3\) is

\[\iint_S f\dd{S} = \iint_D f(\mathbf{r}(u,v))\lvert \mathbf{r}_u \times \mathbf{r}_v\rvert \dd{u}\dd{v}\]

.

Just as the line integral is independent of the curve’s parametrization because it is taken with respect to arc length, the surface integral is independent of the surface’s parametrization because it is taken with respect to the area element. Indeed, if two parametrized surfaces \(\mathbf{r}(u,v)\) and \(\tilde{\mathbf{r}}(s,t)\) have the same image and are related by a \(C^1\) change of variables \((u,v) \mapsto (s,t)\), then by the chain rule \(\mathbf{r}_u \times \mathbf{r}_v = (\partial(s,t)/\partial(u,v))(\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t)\); taking absolute values and applying §Multiple Integrals, ⁋Theorem 4 (Change of variables) shows that the two parametrizations yield the same double integral.

Flux

On the other hand, we may also consider integrating a vector field, rather than a scalar function, over a surface. As in the previous post, we must then decide which side of the surface is “outside.” For a surface, an orientation is a continuous choice of one of the two unit normals \(\pm \mathbf{N}/\lvert \mathbf{N}\rvert\) at each point.

Definition 4 The flux of a continuous vector field \(\mathbf{F}\) across an oriented surface \(S\) with unit normal \(\mathbf{n}\) is

\[\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n}\dd{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)\dd{u}\dd{v}\]

. Here we take \(\mathbf{n} = (\mathbf{r}_u \times \mathbf{r}_v)/\lvert \mathbf{r}_u \times \mathbf{r}_v\rvert\) to match the chosen orientation of the surface.

Flux is the amount of flow across the surface per unit time. For instance, if \(\mathbf{F}\) is the velocity of a fluid, then \(\iint_S \mathbf{F}\cdot d\mathbf{S}\) represents the amount of fluid passing through the surface. It is then intuitively clear that only the normal component \(\mathbf{F}\cdot \mathbf{n}\) contributes to the flow, while the tangential component does not; and one easily checks that reversing the orientation flips \(\mathbf{n}\) and changes the sign of the flux.

Here are two examples of surface integrals.

Example 5 (Surface area of a sphere) Parametrize a sphere of radius \(R\) using spherical coordinates

\[\mathbf{r}(\phi, \theta) = (R\sin\phi\cos\theta, R\sin\phi\sin\theta, R\cos\phi),\qquad 0 \leq \phi \leq \pi,\quad 0 \leq \theta \leq 2\pi\]

. Computing the cross product of the tangent vectors, its magnitude is

\[\lvert \mathbf{r}_\phi \times \mathbf{r}_\theta\rvert = R^2\sin\phi\]

and therefore the surface area is

\[\iint_S \dd{S} = \int_0^{2\pi} \int_0^\pi R^2\sin\phi \dd{\phi} \dd{\theta} = R^2 \cdot 2\pi \cdot 2 = 4\pi R^2\]

which gives the familiar value.

The next example is that of a vector-field integral.

Example 6 On the sphere of Example 5 (Surface area of a sphere), take the outward orientation. Our goal in this example is to compute the flux of

\[\mathbf{F}(x,y,z) = (x,y,z)\]

. As before, in the spherical parametrization,

\[\mathbf{r}_\phi \times \mathbf{r}_\theta = R(\sin\phi) \mathbf{r}\]

and since \(\mathbf{F}(\mathbf{r}) = \mathbf{r}\),

\[\mathbf{F}\cdot(\mathbf{r}_\phi\times \mathbf{r}_\theta) = \mathbf{r} \cdot R\sin\phi \mathbf{r} = R\sin\phi\lvert \mathbf{r}\rvert^2 = R^3\sin\phi\]

. Therefore,

\[\iint_S \mathbf{F}\cdot d\mathbf{S} = \int_0^{2\pi} \int_0^\pi R^3\sin\phi \dd{\phi} \dd{\theta} = 4\pi R^3\]

.

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.