Surface integral

The surface integral or surface integral is a generalization of the integral term on flat or curved surfaces. Thus, the integration domain ${\mathcal {F}}$ is not a one-dimensional interval, but a two-dimensional set in the three-dimensional space $\mathbb {R} ^{3}$ . For a more general representation in $\mathbb {R} ^{n}$ with $n\geq 2$ see: Integration on manifolds.

A general distinction is made between a scalar and a vector surface integral, depending on the form of the integrand and the so-called surface element. They are

Terms and definitions

In integration over surfaces, parameterizations of the surface take the place of the integration variable and surface elements take the place of the infinitesimal (infinitesimal) interval width $\mathrm {d} x$ .

Parameterization

As a two-dimensional set, a surface can be represented (parameterized) as a function of two variables. If $B\subset \mathbb {R} ^{2}$ a set whose boundary contains no double points, is continuously differentiable, is not infinitely long, and further φ $\varphi$ a mapping from into $\mathbb {R} ^{3}$ , we say that φ $\varphi$ is parametrization of the surface ${\mathcal {F}}$ , if ${\mathcal {F}}=\varphi (B)$ . At this point, it is worth noting that much of the difficulty in dealing with surface integrals is related to the parametrization. It is not clear a priori that different parameterizations produce the same value for the integral. A change of coordinates for surface integrals is not trivial and is consequently motivation for the use of differential forms.

general, a surface in $\mathbb {R} ^{3}$ with two parameters and in the following form:

$\varphi \colon B\to \mathbb {R} ^{3},\quad \left(u,v\right)\mapsto {\vec {\varphi }}\left(u,v\right)=\left({\begin{matrix}x\left(u,v\right)\\y\left(u,v\right)\\z\left(u,v\right)\\\end{matrix}}\right)$

On the surface φ the sets of curves $u=\mathrm {const}$ resp. ${\vec {\varphi }}\left(u,v\right)$ form $v=\mathrm {const}$ the coordinate lines. These cover the surface with a coordinate mesh, with two coordinate lines running through each point. Thus, each point on the surface has unique coordinates $\left(u_{0},v_{0}\right)$ .

Example 1: Parameter representation

The surface of a sphere of radius can be parameterized as follows: is the rectangle $[0,\pi ]\times [0,2\pi ]$ and

${\vec {\varphi }}(u,v)={\begin{pmatrix}R\sin(u)\cos(v)\\R\sin(u)\sin(v)\\R\cos(u)\end{pmatrix}}$ .

This parametrization satisfies the spherical equation $x^{2}+y^{2}+z^{2}=R^{2}$ (see also spherical coordinates). is here the polar angle (usually $\vartheta \,$ or θ $\theta \,$ ) and the azimuth angle (usually $\phi \,$ denoted φ {\displaystyle $\varphi \,$ or ϕ ).

Example 2: Explicit representation

If $f\colon B\to \mathbb {R} ,\ \left(x,y\right)\mapsto f\left(x,y\right)$ a function and the surface is z=f(x,y) given in the form , then and are the two parameters; thus the parametrization of the surface looks like this:

${\vec {\varphi }}\left(x,y\right)=\left({\begin{matrix}x\\y\\f\left(x,y\right)\\\end{matrix}}\right)$

Surface element

If in the one-dimensional case the $\mathrm {d} x$ represents the width of an infinitesimal interval, it is natural to replace it in the two-dimensional case by the area of an infinitesimal piece of surface $\mathrm {d} \sigma$ . By the parametrization described in the previous section, one can put two tangents to any point on the surface (see also: curvilinear coordinates): once the tangent that arises if one leaves constant and varies minimally, and once with swapped variables. That is, two tangents to the two coordinate lines at the point under consideration $\left(u_{0},v_{0}\right)$ . These tangents can be expressed by two infinitesimal tangent vectors (let φ ${\vec {\varphi }}\left(u,v\right)$ the parameterized form of the surface):

$\left.{\frac {\partial {\vec {\varphi }}}{\partial u}}\right|_{u_{0},v_{0}}\mathrm {d} u$ and ∂ $\left.{\frac {\partial {\vec {\varphi }}}{\partial v}}\right|_{u_{0},v_{0}}\mathrm {d} v$

In the following, the compact notation is used for the partial derivatives:

${\vec {\varphi }}_{u}={\frac {\partial {\vec {\varphi }}}{\partial u}}$ and φ ${\vec {\varphi }}_{v}={\frac {\partial {\vec {\varphi }}}{\partial v}}$

If these tangents are not parallel at any point on the surface, we speak of a regular parametrization. The cross product of the tangent vectors is then a vector whose length is not equal to zero.

$\left|\left|{\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right|\right|\neq 0$

The two tangent vectors lie in the tangent plane of the surface at the point under consideration. The area of the parallelogram spanned by the two tangent vectors is now just equal to the magnitude of their cross product.

Now if φ is ${\vec {\varphi }}(u,v)$ a regular parametrization of the surface, we define:

Scalar surface element

$\mathrm {d} \sigma =\left|\left|{\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right|\right|\mathrm {d} u\,\mathrm {d} v$

Vector surface element

$\mathrm {d} {\vec {\sigma }}={\hat {n}}\ \mathrm {d} \sigma ={\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\ \mathrm {d} u\,\mathrm {d} v$ with the unit normal vector of the area element ${\hat {n}}={\frac {{\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}}{\left|\left|{\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right|\right|}}$

According to the properties of the cross product, the vector surface element is perpendicular to the surface, its magnitude is just equal to the size of the infinitesimal surface piece.

In the form presented above, the vector surface element is not well-defined, since its direction depends on whether one wants φ ${\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}$ or φ ${\vec {\varphi }}_{v}\times {\vec {\varphi }}_{u}=-\left({\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right)$ computed. The two possibilities are antiparallel to each other. When considering closed surfaces, one usually agrees that the outward facing vector surface element should be used.

Example 1: Parameter representation

The surface of the sphere of radius R can be parameterized by the polar angle and the azimuthal angle as shown above. The surface element is obtained from the following calculation:

${\begin{aligned}&{\vec {\varphi }}=R\left({\begin{matrix}\sin u\ \cos v\\\sin u\ \sin v\\\cos u\\\end{matrix}}\right),\quad {\vec {\varphi }}_{u}=R\left({\begin{matrix}\cos u\ \cos v\\\cos u\ \sin v\\-\sin u\\\end{matrix}}\right),\quad {\vec {\varphi }}_{v}=R\left({\begin{matrix}-\sin u\ \sin v\\\sin u\ \cos v\\0\\\end{matrix}}\right),\\&\pm \left({\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right)=\pm R^{2}\sin u\left({\begin{matrix}\sin u\ \cos v\\\sin u\ \sin v\\\cos u\\\end{matrix}}\right),\quad \left|\left|\pm \left({\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}\right)\right|\right|=R^{2}\sin u,\\&{\hat {n}}=\pm \left({\begin{matrix}\sin u\ \cos v\\\sin u\ \sin v\\\cos u\\\end{matrix}}\right),\quad \mathrm {d} {\vec {\sigma }}={\hat {n}}\,\mathrm {d} \sigma ={\hat {n}}\ R^{2}\sin u\ \mathrm {d} u\,\mathrm {d} v\\\end{aligned}}$

For the normal vector, two solutions are possible ( $\pm$ ), depending on the order of φ ${\vec {\varphi }}_{u}$ and φ ${\vec {\varphi }}_{v}$ in the cross product. Typically, one chooses here the positive solution where ${\hat {n}}$ points away from the convex surface of the sphere (so-called "outer normal").

Example 2: Explicit representation

If the area is of the form z=f(x,y) , then the area element is expressed by the differentials of the coordinates , .

${\vec {\varphi }}=\left({\begin{matrix}x\\y\\f(x,y)\end{matrix}}\right),\quad {\vec {\varphi }}_{x}=\left({\begin{matrix}1\\0\\f_{x}\end{matrix}}\right),\quad {\vec {\varphi }}_{y}=\left({\begin{matrix}0\\1\\f_{y}\end{matrix}}\right)$

$\pm \left({\vec {\varphi }}_{x}\times {\vec {\varphi }}_{y}\right)=\pm \left({\begin{matrix}-f_{x}\\-f_{y}\\1\end{matrix}}\right),\quad \left|\left|\pm \left({\vec {\varphi }}_{x}\times {\vec {\varphi }}_{y}\right)\right|\right|={\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\ ,\quad {\hat {n}}=\pm {\frac {1}{\sqrt {f_{x}^{2}+f_{y}^{2}+1}}}\left({\begin{matrix}-f_{x}\\-f_{y}\\1\end{matrix}}\right)$

Thus, area element and vector area element are the same:

$\mathrm {d} \sigma ={\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\ \mathrm {d} x\,\mathrm {d} y$

$\mathrm {d} {\vec {\sigma }}={\hat {n}}\ {\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\ \mathrm {d} x\,\mathrm {d} y={\begin{pmatrix}-f_{x}\\-f_{y}\\1\end{pmatrix}}\ \mathrm {d} x\,\mathrm {d} y$

Projection on surface with known surface element

We assume in the following that a surface with its surface element $\mathrm {d} A$ and associated normal vector ${\hat {n}}_{A}$ is known. FOR EXAMPLE.

xy-plane:

and $\mathrm {d} A=\mathrm {d} x\mathrm {d} y$ ${\hat {n}}_{A}={\hat {e}}_{z}=\left({\begin{matrix}0\\0\\1\\\end{matrix}}\right)$

Sheath area of a circular cylinder with radius ρ $\rho$ :

and $\mathrm {d} A=\rho \mathrm {d} \varphi \mathrm {d} z$ ${\hat {n}}_{A}={\hat {e}}_{\rho }=\left({\begin{matrix}\cos \varphi \\\sin \varphi \\0\\\end{matrix}}\right)$

Spherical surface with radius :

and $\mathrm {d} A=r^{2}\sin \vartheta \mathrm {d} \vartheta \mathrm {d} \varphi$ ${\hat {n}}_{A}={\hat {e}}_{r}=\left({\begin{matrix}\sin \vartheta \ \cos \varphi \\\sin \vartheta \ \sin \varphi \\\cos \vartheta \\\end{matrix}}\right)$

For another surface ${\mathcal {F}}$ with normal vector ${\hat {n}}_{\mathcal {F}}$ , find the surface element $\mathrm {d} \sigma$ be determined. The area is g(x,y,z)=0 given approximately by and thus the normal vector is equal to ${\hat {n}}_{\mathcal {F}}=\nabla g/||\nabla g||$ .

We now project ${\mathcal {F}}$ along ${\hat {n}}_{A}$ onto . Then the area elements can be calculated using $\mathrm {d} A=\mathrm {d} {\vec {A}}\cdot {\hat {n}}_{A}=|\mathrm {d} {\vec {\sigma }}\cdot {\hat {n}}_{A}|=\mathrm {d} \sigma \,|{\hat {n}}_{\mathcal {F}}\cdot {\hat {n}}_{A}|$ for ${\hat {n}}_{\mathcal {F}}\cdot {\hat {n}}_{A}\neq 0$ link:

$\mathrm {d} \sigma ={\frac {\mathrm {d} A}{|{\hat {n}}_{\mathcal {F}}\cdot {\hat {n}}_{A}|}}={\frac {||\nabla g||\,\mathrm {d} A}{|\nabla g\cdot {\hat {n}}_{A}|}}$

Each straight line along the normal vectors ${\hat {n}}_{A}$ may intersect the surface ${\mathcal {F}}$ only once. Otherwise, one must ${\mathcal {F}}$ divide the surface into smaller surfaces ${\mathcal {F}}_{1},\,{\mathcal {F}}_{2},\,\dotsc$ , whose projection is then unique, or choose a different footprint

The vectorial area element is:

$\mathrm {d} {\vec {\sigma }}={\hat {n}}_{\mathcal {F}}{\frac {\mathrm {d} A}{|{\hat {n}}_{\mathcal {F}}\cdot {\hat {n}}_{A}|}}={\frac {\nabla g}{||\nabla g||}}{\frac {||\nabla g||\,\mathrm {d} A}{|\nabla g\cdot {\hat {n}}_{A}|}}={\frac {\nabla g\,\mathrm {d} A}{|\nabla g\cdot {\hat {n}}_{A}|}}$

Example 1

Let a surface ${\mathcal {F}}$ the form be z=f(x,y) given, then g(x,y,z)=z-f(x,y) and so:

$\nabla g={\begin{pmatrix}-f_{x}\\-f_{y}\\1\end{pmatrix}}\ ,\quad ||\nabla g||={\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\ ,\quad {\hat {n}}_{\mathcal {F}}={\frac {\nabla g}{||\nabla g||}}={\frac {1}{\sqrt {f_{x}^{2}+f_{y}^{2}+1}}}{\begin{pmatrix}-f_{x}\\-f_{y}\\1\end{pmatrix}}$

This surface is now projected into the plane with $\mathrm {d} A=\mathrm {d} x\mathrm {d} y$ and ${\hat {n}}_{A}={\hat {e}}_{z}$ ; where

$\mathrm {d} \sigma ={\frac {||\nabla g||\,\mathrm {d} x\mathrm {d} y}{|\nabla g\cdot {\hat {e}}_{z}|}}={\frac {{\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\,\mathrm {d} x\mathrm {d} y}{|{\hat {e}}_{z}\cdot {\hat {e}}_{z}|}}={\sqrt {f_{x}^{2}+f_{y}^{2}+1}}\,\mathrm {d} x\mathrm {d} y$

$\mathrm {d} {\vec {\sigma }}={\frac {\nabla g\,\mathrm {d} x\mathrm {d} y}{|\nabla g\cdot {\hat {e}}_{z}|}}={\begin{pmatrix}-f_{x}\\-f_{y}\\1\end{pmatrix}}\mathrm {d} x\mathrm {d} y$

Example 2

We are looking for the area element of a body of revolution about the -axis with ρ $\rho =f(z)$ , so $g(\rho ,\varphi ,z)=\rho -f(z)$ .

$\nabla g={\hat {e}}_{\rho }-f_{z}{\hat {e}}_{z}\ ,\quad ||\nabla g||={\sqrt {1+f_{z}^{2}}}\ ,\quad {\hat {n}}_{\mathcal {F}}={\frac {\nabla g}{||\nabla g||}}={\frac {{\hat {e}}_{\rho }-f_{z}{\hat {e}}_{z}}{\sqrt {1+f_{z}^{2}}}}$

By projecting onto the lateral surface of a circular cylinder with radius ρ the area element is obtained: $\rho =f(z)$

$\mathrm {d} \sigma ={\frac {||\nabla g||\,\rho \,\mathrm {d} \varphi \mathrm {d} z}{|\nabla g\cdot {\hat {e}}_{z}|}}={\frac {{\sqrt {1+f_{z}^{2}}}\,f(z)\,\mathrm {d} \varphi \mathrm {d} z}{|({\hat {e}}_{\rho }-f_{z}{\hat {e}}_{z})\cdot {\hat {e}}_{\rho }|}}={\sqrt {1+f_{z}^{2}}}\,f(z)\,\mathrm {d} \varphi \mathrm {d} z$

$\mathrm {d} {\vec {\sigma }}=({\hat {e}}_{\rho }-f_{z}{\hat {e}}_{z})\,f(z)\,\mathrm {d} \varphi \mathrm {d} z$

The integrals

With the parameterizations and the surface elements, one can now define the surface integrals. These multidimensional integrals are Lebesgue integrals, but in most applications they can be calculated as multiple Riemann integrals.

The scalar surface integral

The scalar surface integral of a scalar function $f\colon \mathbb {R} ^{3}\rightarrow \mathbb {R}$ over a surface ${\mathcal {F}}$ with regular parametrization φ $\varphi \colon B\rightarrow \mathbb {R} ^{3}$ with $B\subset \mathbb {R} ^{2}$ is defined as

$\iint _{\mathcal {F}}f({\vec {x}})\,\mathrm {d} \sigma =\iint _{B}f\left({\vec {\varphi }}(u,v)\right)\,||{\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v}||\,\,\mathrm {d} (u,v)$ .

For example, if we set $f({\vec {x}})=1$ , then the scalar surface integral is simply the area of the surface.

Example: Surface area of a sphere

With the area element for spherical coordinates

$\mathrm {d} \sigma =r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi$

obtained for the surface area ${\mathcal {F}}$ a sphere of radius :

${\mathcal {F}}=\iint _{\mathcal {F}}1\,\mathrm {d} \sigma =\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi =r^{2}\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }\sin \theta \,\mathrm {d} \theta \ \mathrm {d} \varphi =r^{2}\int \limits _{0}^{2\pi }2\,\mathrm {d} \varphi =4\pi r^{2}$ .

The vectorial surface integral

The vector surface integral of a vector-valued function $f\colon \mathbb {R} ^{3}\rightarrow \mathbb {R} ^{3}$ over a surface ${\mathcal {F}}$ with regular parametrization φ $\varphi \colon B\rightarrow \mathbb {R} ^{3}$ with $B\subset \mathbb {R} ^{2}$ is defined as

$\iint _{\mathcal {F}}{\vec {f}}({\vec {x}})\cdot \mathrm {d} {\vec {\sigma }}=\iint _{B}{\vec {f}}\left({\vec {\varphi }}(u,v)\right)\cdot ({\vec {\varphi }}_{u}\times {\vec {\varphi }}_{v})\,\,\mathrm {d} (u,v)=:\Phi _{\mathcal {F}}({\vec {f}})$ .

A descriptive notion of this integral occurs via the flow $\Phi$ a vector field ${\vec {f}}$ through the surface ${\mathcal {F}}$ : The quantity ${\vec {f}}\cdot \mathrm {d} {\vec {\sigma }}$ indicates, what contribution to the total flux $\Phi _{\mathcal {F}}({\vec {f}})$ the infinitesimal-small surface vector $\mathrm {d} {\vec {\sigma }}={\hat {n}}\,\mathrm {d} \sigma$ provides; namely, how much of $\mathrm {d} {\sigma }$ flows ${\vec {f}}$ through the surface piece The flow is maximal when the vector field ${\vec {f}}$ parallel to the surface normal ${\hat {n}}$ and zero, if ${\vec {f}}$ perpendicular to ${\hat {n}}$ that is, tangent to the surface - then "flows" ${\vec {f}}$ along the surface, but not through it.

Example: Flow of a vector field through a spherical surface

Given a radially symmetric vector field

${\vec {E}}({\vec {r}})={\dfrac {C}{r^{2}}}\cdot {\dfrac {\vec {r}}{r}}$

With a constant $C\in \mathbb{R}$ , the location vector ${\vec {r}}$ and its magnitude . The vector ${\dfrac {\vec {r}}{r}}$ is thus a unit vector in the direction of the location vector. In physics, for example, the electric field of a point charge at the coordinate origin is of this form: see Coulomb's law.

For symmetry reasons one uses spherical coordinates. The vector surface element $\textstyle \mathrm {d} {\vec {\sigma }}$ for a sphere with radius and center at the coordinate origin is

$\mathrm {d} {\vec {\sigma }}=r^{2}\sin \theta \cdot {\dfrac {\vec {r}}{r}}\,\mathrm {d} \theta \,\mathrm {d} \varphi$ .

For the flux $\Phi$ of the vector field ${\vec {E}}({\vec {r}})$ through the surface $\mathcal F$ a sphere of radius obtain:

$\Phi =\iint _{\mathcal {F}}{\dfrac {C}{r^{2}}}\cdot {\dfrac {\vec {r}}{r}}\,\mathrm {d} {\vec {\sigma }}=\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }{\dfrac {C}{r^{2}}}{\dfrac {\vec {r}}{r}}\cdot r^{2}\sin \theta \cdot {\dfrac {\vec {r}}{r}}\,\mathrm {d} \theta \,\mathrm {d} \varphi =C\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi =C\int \limits _{0}^{2\pi }2\mathrm {d} \varphi =4\pi C$ .

The flux $\Phi$ the vector field through the sphere surface is thus independent of the sphere radius . For the physical example of the electric field of a point charge, this result is a special case of Gauss' law of electrostatics.

Questions and Answers

Q: What is a surface integral in mathematics?

A: A surface integral is a definite integral taken over a surface, which may be a curve set in space.

Q: What is the difference between a line integral and a surface integral?

A: A line integral allows one to integrate over an arbitrary curve of one dimension, while a surface integral can be thought of as a double integral integrating over a two-dimensional surface.

Q: What can be integrated over a surface?

A: Given a surface, one may integrate over its scalar fields or its vector fields.

Q: What are scalar fields?

A: Scalar fields are functions which return numbers as values.

Q: What are vector fields?

A: Vector fields are functions which return vectors as values.

Q: What are the applications of surface integrals in physics?

A: Surface integrals have applications in physics, particularly with the classical theory of electromagnetism.

Q: How can surface integrals be used in electromagnetism?

A: Surface integrals are useful in calculating magnetic flux or electric flux through a closed surface.

Surface integral

Terms and definitions

Parameterization

Example 1: Parameter representation

Example 2: Explicit representation

Surface element

Example 1: Parameter representation

Example 2: Explicit representation

Projection on surface with known surface element

Example 1

Example 2

The integrals

The scalar surface integral

The vectorial surface integral

Questions and Answers

Q: What is a surface integral in mathematics?

Q: What is the difference between a line integral and a surface integral?

Q: What can be integrated over a surface?

Q: What are scalar fields?

Q: What are vector fields?

Q: What are the applications of surface integrals in physics?

Q: How can surface integrals be used in electromagnetism?

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