- Surface integral
In

mathematics , a**surface integral**is adefinite integral taken over asurface (which may be a curved set inspace ); it can be thought of as thedouble integral analog of theline integral . Given a surface, one may integrate over itscalar field s (that is, functions which returnnumber s as values), andvector field s (that is, functions which return vectors as values).Surface integrals have applications in

physics , particularly with theclassical theory ofelectromagnetism .**Surface integrals of scalar fields**Consider a surface "S" on which a scalar field "f" is defined. If we think of "S" as made of some material, and for each

**x**in "S" the number "f"(**x**) is thedensity of material at**x**, then the surface integral of "f" over "S" is themass per unit thickness of "S". (This only holds as long as the surface is an infinitesimally thin shell.) One approach to calculate the surface integral is then to split the surface in many very small pieces, assume that on each piece the density is approximately constant, find the mass per unit thickness of each piece by multiplying the density of the piece by its area, and then sum up the resulting numbers to find the total mass per unit thickness of "S".To find an explicit formula for the surface integral, we need to parametrize "S" by considering on "S" a system of

curvilinear coordinates , like the latitude and longitude on asphere . Let such a parametrization be**x**("s", "t"), where ("s", "t") varies in some region "T" in the plane. Then, the surface integral is given by:$int\_S\; f\; ,dS\; =\; iint\_T\; f(mathbf\{x\}(s,\; t))\; left|\{partial\; mathbf\{x\}\; over\; partial\; s\}\; imes\; \{partial\; mathbf\{x\}\; over\; partial\; t\}\; ight|\; ds,\; dt$ where the expression between bars on the right-hand side is the magnitude of the

cross product of thepartial derivative s of**x**("s", "t").For example, if we want to find the surface area of some general functional shape, say $z=f,(x,y)$, we have:$A\; =\; int\_S\; ,dS\; =\; iint\_T\; left|\{partial\; mathbf\{r\}\; over\; partial\; x\}\; imes\; \{partial\; mathbf\{r\}\; over\; partial\; y\}\; ight|\; dx,\; dy$where $mathbf\{r\}=(x,\; y,\; z)=(x,\; y,\; f(x,y))$. So, $\{partial\; mathbf\{r\}\; over\; partial\; x\}=(1,\; 0,\; f\_x(x,y))$, and $\{partial\; mathbf\{r\}\; over\; partial\; y\}=(0,\; 1,\; f\_y(x,y))$. So, :$egin\{align\}A\; \{\}\; =\; iint\_T\; left|left(1,\; 0,\; \{partial\; f\; over\; partial\; x\}\; ight)\; imes\; left(0,\; 1,\; \{partial\; f\; over\; partial\; y\}\; ight)\; ight|\; dx,\; dy\; \backslash \{\}\; =\; iint\_T\; left|left(-\{partial\; f\; over\; partial\; x\},\; -\{partial\; f\; over\; partial\; y\},\; 1\; ight)\; ight|\; dx,\; dy\; \backslash \{\}\; =\; iint\_T\; sqrt\{left(\{partial\; f\; over\; partial\; x\}\; ight)^2+left(\{partial\; f\; over\; partial\; y\}\; ight)^2+1\},\; ,\; dx,\; dyend\{align\}$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.

**Surface integrals of vector fields**Consider a vector field

**v**on "S", that is, for each**x**in "S",**v**(**x**) is a vector. Imagine that we have a fluid flowing through "S", such that**v**(**x**) determines the velocity of the fluid at**x**. Theflux is defined as the quantity of fluid flowing through "S" in unit amount of 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 anormal component , then only the normal component contributes to the flux. Based on this reasoning, to find the flux, we need to take thedot product of**v**with the unitsurface normal to "S" at each point, which will give us a scalar field, and integrate the obtained field as above. We find the formula:$int\_S\; \{mathbf\; v\}cdot\; ,d\{mathbf\; \{S\; =\; int\_S\; (\{mathbf\; v\}cdot\; \{mathbf\; n\}),dS=iint\_T\; \{mathbf\; v\}(mathbf\{x\}(s,\; t))cdot\; left(\{partial\; mathbf\{x\}\; over\; partial\; s\}\; imes\; \{partial\; mathbf\{x\}\; over\; partial\; t\}\; ight)\; ds,\; dt.$

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

This formula is "defined" to be the integral of the vector field

**v**on "S".**Surface integrals of differential 2-forms**Let :$f=f\_\{1\}\; dx\; wedge\; dy\; +\; f\_\{2\}\; dy\; wedge\; dz\; +\; f\_\{3\}\; dz\; wedge\; dx$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". Then, the surface integral of "f" on "S" is given by

:$iint\_D\; left\; [\; f\_\{1\}\; (\; mathbf\{x\}\; (s,t))\; frac\{partial(x,y)\}\{partial(s,t)\}\; +\; f\_\{2\}\; (\; mathbf\{x\}\; (s,t))frac\{partial(y,z)\}\{partial(s,t)\}\; +\; f\_\{3\}\; (\; mathbf\{x\}\; (s,t))frac\{partial(z,x)\}\{partial(s,t)\}\; ight]\; ,\; ds\; dt$

where :$\{partial\; mathbf\{x\}\; over\; partial\; s\}\; imes\; \{partial\; mathbf\{x\}\; over\; partial\; t\}=left(frac\{partial(y,z)\}\{partial(s,t)\},\; frac\{partial(z,x)\}\{partial(s,t)\},\; frac\{partial(x,y)\}\{partial(s,t))\}\; ight)$is the surface 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 $f\_1$, $f\_2$ and $f\_3.$

**Theorems involving surface integrals**Various useful results for surface integrals can be derived using

differential geometry andvector calculus , such as thedivergence theorem , and its generalization,Stokes' theorem .**Advanced issues**Let us notice that we defined the surface integral by using a parametrization of the surface "S". We know that a given surface might have several parametrizations. For example, if we move the locations of the North Pole and 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 proved 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 which direction the normal will point to and then choose any parametrization consistent with that direction.

Another issue is that sometimes surfaces do not have parametrizations which cover the whole surface; this is true for example for the surface of a cylinder (of finite height). The obvious solution is then to split that surface in 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.

Lastly, there are surfaces which do not admit a surface normal at each point with consistent results (for example, the

Möbius 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 called non-orientable, and on this kind of surface one cannot talk about integrating vector fields.**ee also***

Divergence theorem

*Stokes' theorem

*Volume integral

*Volume and surface elements in different co-ordinate systems **External links*** [

*http://mathworld.wolfram.com/SurfaceIntegral.html Surface Integral -- from MathWorld*]

* [*http://www.math.gatech.edu/%7Ecain/notes/cal15.pdf Surface Integral -- Theory and exercises*]

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### Look at other dictionaries:

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