# Surface integral

In mathematics, a surface integral is a generalization of multiple integrals to integration over surfaces. It can be thought of as the double integral analog of the line integral. Given a surface, one may integrate over its scalar fields (that is, functions which return scalars as values), and vector fields (that is, functions which return vectors as values).

Surface integrals have applications in physics, particularly with the theories of classical electromagnetism.

The definition of surface integral relies on splitting the surface into small surface elements.
An illustration of a single surface element. These elements are made infinitesimally small, by the limiting process, so as to approximate the surface.

## Surface integrals of scalar fields

To find an explicit formula for the surface integral, we need to parameterize the surface of interest, S, by considering a system of curvilinear coordinates on S, like the latitude and longitude on a sphere. Let such a parameterization be x(s, t), where (s, t) varies in some region T in the plane. Then, the surface integral is given by

$\iint_{S} f \,\mathrm dS = \iint_{T} f(\mathbf{x}(s, t)) \left\|{\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right\| \mathrm ds\, \mathrm dt$

where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of x(s, t), and is known as the surface element.

For example, if we want to find the surface area of some general scalar function, say $z=f\,(x,y)$, we have

$A = \iint_S \,\mathrm dS = \iint_T \left\|{\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y}\right\| \mathrm dx\, \mathrm dy$

where $\mathbf{r}=(x, y, z)=(x, y, f(x,y))$. So that ${\partial \mathbf{r} \over \partial x}=(1, 0, f_x(x,y))$, and ${\partial \mathbf{r} \over \partial y}=(0, 1, f_y(x,y))$. So,

\begin{align} A &{} = \iint_T \left\|\left(1, 0, {\partial f \over \partial x}\right)\times \left(0, 1, {\partial f \over \partial y}\right)\right\| \mathrm dx\, \mathrm dy \\ &{} = \iint_T \left\|\left(-{\partial f \over \partial x}, -{\partial f \over \partial y}, 1\right)\right\| \mathrm dx\, \mathrm dy \\ &{} = \iint_T \sqrt{\left({\partial f \over \partial x}\right)^2+\left({\partial f \over \partial y}\right)^2+1}\, \, \mathrm dx\, \mathrm dy \end{align}

which is the familiar formula we get for the surface area of a general functional shape. One can recognize the vector in the second line above as the normal vector to the surface.

Note that because of the presence of the cross product, the above formulas only work for surfaces embedded in three-dimensional space.

This can be seen as integrating a Riemannian volume form on the parameterized surface, where the metric tensor is given by the first fundamental form of the surface.

## Surface integrals of vector fields

A vector field on a surface

Consider a vector field v on S, that is, for each x in S, v(x) is a vector.

The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. This applies for example in the expression of the electric field at some fixed point due to an electrically charged surface, or the gravity at some fixed point due to a sheet of material.

Alternatively, if we integrate the normal component of the vector field, the result is a scalar. Imagine that we have a fluid flowing through S, such that v(x) determines the velocity of the fluid at x. The flux is defined as the quantity of fluid flowing through S per unit time.

This illustration implies that if the vector field is tangent to S at each point, then the flux is zero, because the fluid just flows in parallel to S, and neither in nor out. This also implies that if v does not just flow along S, that is, if v has both a tangential and a normal component, then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take the dot product of v with the unit surface normal to S at each point, which will give us a scalar field, and integrate the obtained field as above. We find the formula

$\iint_S {\mathbf v}\cdot\mathrm d{\mathbf {S}} = \iint_S {\mathbf v}\cdot {\mathbf n}\,\mathrm dS=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \mathrm ds\, \mathrm dt.$

The cross product on the right-hand side of this expression is a surface normal determined by the parametrization.

This formula defines the integral on the left (note the dot and the vector notation for the surface element).

We may also interpret this as a special case of integrating 2-forms, where we identify the vector field with a 1-form, and then integrate its Hodge dual over the surface. This is equivalent to integrating $\langle \mathbf{v}, \mathbf{n} \rangle \;\mathrm dS$ over the immersed surface, where $\mathrm dS$ is the induced volume form on the surface, obtained by interior multiplication of the Riemannian metric of the ambient space with the outward normal of the surface.

## Surface integrals of differential 2-forms

Let

$f=f_{z}\, \mathrm dx \wedge \mathrm dy + f_{x}\, \mathrm dy \wedge \mathrm dz + f_{y}\, \mathrm dz \wedge \mathrm dx$

be a differential 2-form defined on the surface S, and let

$\mathbf{x} (s,t)=( x(s,t), y(s,t), z(s,t))\!$

be an orientation preserving parametrization of S with $(s,t)$ in D. Changing coordinates from $(x, y)$ to $(s, t)$, the differential forms transform as

$\mathrm dx=\frac{\mathrm dx}{\mathrm ds}\mathrm ds+\frac{\mathrm dx}{\mathrm dt}\mathrm dt$
$\mathrm dy=\frac{\mathrm dy}{\mathrm ds}\mathrm ds+\frac{\mathrm dy}{\mathrm dt}\mathrm dt$

So $\mathrm dx \wedge \mathrm dy$ transforms to $\frac{\partial(x,y)}{\partial(s,t)} \mathrm ds \wedge \mathrm dt$, where $\frac{\partial(x,y)}{\partial(s,t)}$ denotes the determinant of the Jacobian of the transition function from $(s, t)$ to $(x,y)$. The transformation of the other forms are similar.

Then, the surface integral of f on S is given by

$\iint_D \left[ f_{z} ( \mathbf{x} (s,t)) \frac{\partial(x,y)}{\partial(s,t)} + f_{x} ( \mathbf{x} (s,t))\frac{\partial(y,z)}{\partial(s,t)} + f_{y} ( \mathbf{x} (s,t))\frac{\partial(z,x)}{\partial(s,t)} \right]\, \mathrm ds\, \mathrm dt$

where

${\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}=\left(\frac{\partial(y,z)}{\partial(s,t)}, \frac{\partial(z,x)}{\partial(s,t)}, \frac{\partial(x,y)}{\partial(s,t)}\right)$

is the surface element normal to S.

Let us note that the surface integral of this 2-form is the same as the surface integral of the vector field which has as components $f_x$, $f_y$ and $f_z$.

## Theorems involving surface integrals

Various useful results for surface integrals can be derived using differential geometry and vector calculus, such as the divergence theorem, and its generalization, Stokes' theorem.