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Surface integral

From Wikipedia, the free encyclopedia
Integration over a non-flat region in 3D space
The definition of the surface integral relies on splitting the surface into small surface elements.
Part of a series of articles about
Calculus
∫abf′(t)dt=f(b)−f(a){\displaystyle \int _{a}^{b}f'(t)\,dt=f(b)-f(a)}

Inmathematics, particularlymultivariable calculus, asurface integral is a generalization ofmultiple integrals tointegration oversurfaces. It can be thought of as thedouble integral analogue of theline integral. Given a surface, one may integrate over this surface ascalar field (that is, afunction of position which returns ascalar as a value), or avector field (that is, a function which returns avector as value). If a region R is not flat, then it is called asurface as shown in the illustration.

Surface integrals have applications inphysics, particularly in theclassical theories ofelectromagnetism andfluid mechanics.

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

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Assume thatf is a scalar, vector, or tensor field defined on a surfaceS.To find an explicit formula for the surface integral off overS, we need toparameterizeS by defining a system ofcurvilinear coordinates onS, like thelatitude and longitude on asphere. Let such a parameterization ber(s,t), where(s,t) varies in some regionT in theplane. Then, the surface integral is given by

∬SfdS=∬Tf(r(s,t))‖∂r∂s×∂r∂t‖dsdt{\displaystyle \iint _{S}f\,\mathrm {d} S=\iint _{T}f(\mathbf {r} (s,t))\left\|{\partial \mathbf {r} \over \partial s}\times {\partial \mathbf {r} \over \partial t}\right\|\mathrm {d} s\,\mathrm {d} t}

where the expression between bars on the right-hand side is themagnitude of thecross product of thepartial derivatives ofr(s,t), and is known as the surfaceelement (which would, for example, yield a smaller value near the poles of a sphere, where the lines of longitude converge more dramatically, and latitudinal coordinates are more compactly spaced). The surface integral can also be expressed in the equivalent form

∬SfdS=∬Tf(r(s,t))gdsdt{\displaystyle \iint _{S}f\,\mathrm {d} S=\iint _{T}f(\mathbf {r} (s,t)){\sqrt {g}}\,\mathrm {d} s\,\mathrm {d} t}

whereg is the determinant of thefirst fundamental form of the surface mappingr(s,t).[1][2]

For example, if we want to find thesurface area of the graph of some scalar function, sayz =f(x,y), we have

A=∬SdS=∬T‖∂r∂x×∂r∂y‖dxdy{\displaystyle A=\iint _{S}\,\mathrm {d} S=\iint _{T}\left\|{\partial \mathbf {r} \over \partial x}\times {\partial \mathbf {r} \over \partial y}\right\|\mathrm {d} x\,\mathrm {d} y}

wherer = (x,y,z) = (x,y,f(x,y)). So that∂r∂x=(1,0,fx(x,y)){\displaystyle {\partial \mathbf {r} \over \partial x}=(1,0,f_{x}(x,y))}, and∂r∂y=(0,1,fy(x,y)){\displaystyle {\partial \mathbf {r} \over \partial y}=(0,1,f_{y}(x,y))}. So,

A=∬T‖(1,0,∂f∂x)×(0,1,∂f∂y)‖dxdy=∬T‖(−∂f∂x,−∂f∂y,1)‖dxdy=∬T(∂f∂x)2+(∂f∂y)2+1dxdy{\displaystyle {\begin{aligned}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 {d} x\,\mathrm {d} y\\&{}=\iint _{T}\left\|\left(-{\partial f \over \partial x},-{\partial f \over \partial y},1\right)\right\|\mathrm {d} x\,\mathrm {d} y\\&{}=\iint _{T}{\sqrt {\left({\partial f \over \partial x}\right)^{2}+\left({\partial f \over \partial y}\right)^{2}+1}}\,\,\mathrm {d} x\,\mathrm {d} y\end{aligned}}}

which is the standard formula for the area of a surface described this way. One can recognize the vector in the second-last line above as thenormal vector to the surface.

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 aRiemannian volume form on the parameterized surface, where themetric tensor is given by thefirst fundamental form of the surface.

Surface integrals of vector fields

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A curved surfaceS{\displaystyle S} with a vector fieldF{\displaystyle \mathbf {F} } passing through it. The red arrows (vectors) represent the magnitude and direction of the field at various points on the surface
Surface divided into small patchesdS=dudv{\displaystyle dS=du\,dv} by a parameterization of the surface[u(x),v(x)]{\displaystyle [u(\mathbf {x} ),v(\mathbf {x} )]}
The flux through each patch is equal to the normal (perpendicular) component of the fieldFn(x)=F(x)cos⁡θ{\displaystyle F_{n}(\mathbf {x} )=F(\mathbf {x} )\cos \theta } at the patch's locationx{\displaystyle \mathbf {x} } multiplied by the areadS{\displaystyle dS}. The normal component is equal to thedot product ofF(x){\displaystyle \mathbf {F} (\mathbf {x} )} with the unit normal vectorn(x){\displaystyle \mathbf {n} (\mathbf {x} )}(blue arrows)
The total flux through the surface is found by adding upF⋅ndS{\displaystyle \mathbf {F} \cdot \mathbf {n} \;dS} for each patch. In the limit as the patches become infinitesimally small, this is the surface integral
∬SF⋅ndS{\textstyle \iint _{S}\mathbf {F\cdot n} \;dS}

Consider a vector fieldv on a surfaceS, that is, for eachr = (x,y,z) inS,v(r) is a vector.

The integral ofv onS was defined in the previous section. Suppose now that it is desired to integrate onlythenormal component of the vector field over the surface, the result being a scalar, usually called theflux passing through the surface. For example, imagine that we have a fluid flowing throughS, such thatv(r) determines the velocity of the fluid atr. Theflux is defined as the quantity of fluid flowing throughS per unit time.

This illustration implies that if the vector field istangent toS at each point, then the flux is zero because the fluid just flows inparallel toS, and neither in nor out. This also implies that ifv does not just flow alongS, that is, ifv 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 thedot product ofv with the unitsurface normaln toS at each point, which will give us a scalar field, and integrate the obtained field as above. In other words, we have to integratev with respect to the vector surface elementds=nds{\displaystyle \mathrm {d} \mathbf {s} ={\mathbf {n} }\mathrm {d} s}, which is the vector normal toS at the given point, whose magnitude isds=‖ds‖.{\displaystyle \mathrm {d} s=\|\mathrm {d} {\mathbf {s} }\|.}

We find the formula

∬Sv⋅ds=∬S(v⋅n)ds=∬T(v(r(s,t))⋅∂r∂s×∂r∂t‖∂r∂s×∂r∂t‖)‖∂r∂s×∂r∂t‖dsdt=∬Tv(r(s,t))⋅(∂r∂s×∂r∂t)dsdt.{\displaystyle {\begin{aligned}\iint _{S}{\mathbf {v} }\cdot \mathrm {d} {\mathbf {s} }&=\iint _{S}\left({\mathbf {v} }\cdot {\mathbf {n} }\right)\,\mathrm {d} s\\&{}=\iint _{T}\left({\mathbf {v} }(\mathbf {r} (s,t))\cdot {{\frac {\partial \mathbf {r} }{\partial s}}\times {\frac {\partial \mathbf {r} }{\partial t}} \over \left\|{\frac {\partial \mathbf {r} }{\partial s}}\times {\frac {\partial \mathbf {r} }{\partial t}}\right\|}\right)\left\|{\frac {\partial \mathbf {r} }{\partial s}}\times {\frac {\partial \mathbf {r} }{\partial t}}\right\|\mathrm {d} s\,\mathrm {d} t\\&{}=\iint _{T}{\mathbf {v} }(\mathbf {r} (s,t))\cdot \left({\frac {\partial \mathbf {r} }{\partial s}}\times {\frac {\partial \mathbf {r} }{\partial t}}\right)\mathrm {d} s\,\mathrm {d} t.\end{aligned}}}

The cross product on the right-hand side of this expression is a (not necessarily unital) surface normal determined by the parametrisation.

This formuladefines 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 itsHodge dual over the surface.This is equivalent to integrating⟨v,n⟩dS{\displaystyle \left\langle \mathbf {v} ,\mathbf {n} \right\rangle \mathrm {d} S} over the immersed surface, wheredS{\displaystyle \mathrm {d} S} is the induced volume form on the surface, obtainedbyinterior multiplication of the Riemannian metric of the ambient space with the outward normal of the surface.

Surface integrals of differential 2-forms

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Let

f=dxdyfz+dydzfx+dzdxfy{\displaystyle f=\mathrm {d} x\mathrm {d} y\,f_{z}+\mathrm {d} y\mathrm {d} z\,f_{x}+\mathrm {d} z\mathrm {d} x\,f_{y}}

be adifferential 2-form defined on a surfaceS, and let

r(s,t)=(x(s,t),y(s,t),z(s,t)){\displaystyle \mathbf {r} (s,t)=(x(s,t),y(s,t),z(s,t))}

be anorientation preserving parametrization ofS with(s,t){\displaystyle (s,t)} inD. Changing coordinates from(x,y){\displaystyle (x,y)} to(s,t){\displaystyle (s,t)}, the differential forms transform as

dx=∂x∂sds+∂x∂tdt{\displaystyle \mathrm {d} x={\frac {\partial x}{\partial s}}\mathrm {d} s+{\frac {\partial x}{\partial t}}\mathrm {d} t}
dy=∂y∂sds+∂y∂tdt{\displaystyle \mathrm {d} y={\frac {\partial y}{\partial s}}\mathrm {d} s+{\frac {\partial y}{\partial t}}\mathrm {d} t}

Sodxdy{\displaystyle \mathrm {d} x\mathrm {d} y} transforms to∂(x,y)∂(s,t)dsdt{\displaystyle {\frac {\partial (x,y)}{\partial (s,t)}}\mathrm {d} s\mathrm {d} t}, where∂(x,y)∂(s,t){\displaystyle {\frac {\partial (x,y)}{\partial (s,t)}}} denotes thedeterminant of theJacobian of the transition function from(s,t){\displaystyle (s,t)} to(x,y){\displaystyle (x,y)}. The transformation of the other forms are similar.

Then, the surface integral off onS is given by

∬D[fz(r(s,t))∂(x,y)∂(s,t)+fx(r(s,t))∂(y,z)∂(s,t)+fy(r(s,t))∂(z,x)∂(s,t)]dsdt{\displaystyle \iint _{D}\left[f_{z}(\mathbf {r} (s,t)){\frac {\partial (x,y)}{\partial (s,t)}}+f_{x}(\mathbf {r} (s,t)){\frac {\partial (y,z)}{\partial (s,t)}}+f_{y}(\mathbf {r} (s,t)){\frac {\partial (z,x)}{\partial (s,t)}}\right]\,\mathrm {d} s\,\mathrm {d} t}

where

∂r∂s×∂r∂t=(∂(y,z)∂(s,t),∂(z,x)∂(s,t),∂(x,y)∂(s,t)){\displaystyle {\partial \mathbf {r} \over \partial s}\times {\partial \mathbf {r} \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 toS.

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 componentsfx{\displaystyle f_{x}},fy{\displaystyle f_{y}} andfz{\displaystyle f_{z}}.

Theorems involving surface integrals

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Various useful results for surface integrals can be derived usingdifferential geometry andvector calculus, such as thedivergence theorem,magnetic flux, and its generalization,Stokes' theorem.

Dependence on parametrization

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Let us notice that we defined the surface integral by using a parametrization of the surfaceS. We know that a given surface might have several parametrizations. For example, if we move the locations of the North Pole and the South Pole on a sphere, the latitude and longitude change for all the points on the sphere. A natural question is then whether the definition of the surface integral depends on the chosen parametrization. For integrals of scalar fields, the answer to this question is simple; the value of the surface integral will be the same no matter what parametrization one uses.

For integrals of vector fields, things are more complicated because the surface normal is involved. It can be proven that given two parametrizations of the same surface, whose surface normals point in the same direction, one obtains the same value for the surface integral with both parametrizations. If, however, the normals for these parametrizations point in opposite directions, the value of the surface integral obtained using one parametrization is the negative of the one obtained via the other parametrization. It follows that given a surface, we do not need to stick to any unique parametrization, but, when integrating vector fields, we do need to decide in advance in which direction the normal will point and then choose any parametrization consistent with that direction.

Another issue is that sometimes surfaces do not have parametrizations which cover the whole surface. The obvious solution is then to split that surface into several pieces, calculate the surface integral on each piece, and then add them all up. This is indeed how things work, but when integrating vector fields, one needs to again be careful how to choose the normal-pointing vector for each piece of the surface, so that when the pieces are put back together, the results are consistent. For the cylinder, this means that if we decide that for the side region the normal will point out of the body, then for the top and bottom circular parts, the normal must point out of the body too.

Last, there are surfaces which do not admit a surface normal at each point with consistent results (for example, theMöbius strip). If such a surface is split into pieces, on each piece a parametrization and corresponding surface normal is chosen, and the pieces are put back together, we will find that the normal vectors coming from different pieces cannot be reconciled. This means that at some junction between two pieces we will have normal vectors pointing in opposite directions. Such a surface is callednon-orientable, and on this kind of surface, one cannot talk about integrating vector fields.

See also

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References

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  1. ^Edwards, C. H. (1994).Advanced Calculus of Several Variables. Mineola, NY: Dover. p. 335.ISBN 0-486-68336-2.
  2. ^Hazewinkel, Michiel (2001) [1994]."Surface Integral".Encyclopedia of Mathematics.EMS Press.

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