# Surface Integral

In mathematics , especially multivariate calculus , a surface integral is a generalization of the integration to many integrals over surfaces . It can be thought of as the double integral analog of the integral line . Given a surface, one can integrate a scalar field (that is, a function of position that returns a scalar as a value) over the surface, or a vector field (that is, a function that returns a vector as a value ). returns as). If an area R is not flat, it is called a surface .As shown in the illustration.

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

## Surface integration of scalar fields

To find an explicit formula for the surface integral S on a surface , we need to parameterize S by defining a system of curvilinear coordinates at S , such as latitude and longitude on an area . Let such a parameterization be r ( s , t ) , where ( s , t ) varies in some region T of the plane . Then, the page integration is given by

{\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 the bars on the right is the magnitude of the cross product of the partial derivatives of r ( s , t ) , and is referred to as the surface element . Surface integration can also be expressed as

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


where g is the determinant of the first fundamental form of the surface mapping r ( s , t ) . [1] [2] For example, if we want to find the surface area of ​​the graph of a scalar function, such as z = f ( x , y ) , we have

{\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}

where r = ( x , y , z ) = ( x , y , f ( x , y )) . Because , and . therefore,

,{\partial \mathbf{r} \over \partial x}=(1, 0, f_x(x,y))
{\partial \mathbf{r} \over \partial y}=(0, 1, f_y(x,y))
\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 standard formula for the area of ​​a surface described as such. The vector in the second-last row above can be recognized as a vector normal to the surface .

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

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

## Surface Integration of Vector Fields

Consider a vector field v on a surface S , that is, for each r = ( x , y , z ) in s , v ( r ) is a vector.

The surface integration 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 to the expression of an electric field at a certain point due to an electrically charged surface, or to gravity at a certain point due to a sheet of material.

Alternatively, if we integrate the normal component of the vector field over the surface , the result is a scalar, usually called the flux passing through the surface . Imagine that we have a fluid flowing through S , such that v ( r ) determines the velocity of the fluid at r . Flow is defined as the amount of fluid flowing through s per unit time.

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

{\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 unitary) surface normal determined by parametrization.

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

We can also interpret this as a special case of integrating 2-form, where we identify the vector field with 1-form, and then integrate its Hodge dual on the surface. This is equivalent to integrating on the immersed surface, where the induced volume at the surface is of the form, obtained by the internal multiplication of the Riemannian metric of the ambient space with the external normal of the surface.

## Surface Integration of Differential 2-Forms

Army

{\displaystyle f=f_{z}\,\mathrm {d} x\wedge \mathrm {d} y+f_{x}\,\mathrm {d} y\wedge \mathrm {d} z+f_{y}\,\mathrm {d} z\wedge \mathrm {d} x}

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

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

be an orientation preserving parametrization with s(s , t)in D. _ Converting from coordinates to , differential form transforms(x , y)(s , t)

{\displaystyle \mathrm {d} x={\frac {\partial x}{\partial s}}\mathrm {d} s+{\frac {\partial x}{\partial t}}\mathrm {d} t}
{\displaystyle \mathrm {d} y={\frac {\partial y}{\partial s}}\mathrm {d} s+{\frac {\partial y}{\partial t}}\mathrm {d} t}

So in turns , where is the meaning of the determinant of the Jacobian transition function to . The transformation of other forms is similar.

\mathrm dx \wedge \mathrm dy
\frac{\partial(x,y)}{\partial(s,t)} \mathrm ds \wedge \mathrm dt\frac{\partial(x,y)}{\partial(s,t)}(s, t)(x,y)

Then, the surface integration of f at S is given by?

{\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 from

{\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)}

The surface element common to S is .

Let us note that the surface integration of this 2-form is the same as the surface integration of a vector field that has , and .

f_{x}f_yf_z


## Surface Integration Theorems

Various useful results for surface integration can be obtained using differential geometry and vector calculus, such as the divergence theorem, and its generalization, Stokes’ theorem.

## Dependency on parameterization

Let’s note that we have defined the surface integral using the parametrization of the surface S. We know that a given surface can have multiple parametrizations. For example, if we rotate the North Pole and South Pole locations on a sphere, the latitude and longitude of all the points on the sphere change. A natural question is whether the definition of the surface integral depends on the chosen parametrization. For integration of scalar fields, the answer to this question is simple; The value of the surface integral will be the same whatever parametrization is used.

For integrals of vector fields, things are more complicated because surface normals are 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 integral of the surface with both parametrizations. If, however, the normals for these parametrizations point in opposite directions, then the value of the surface integral obtained using one parametrization is negative of the one obtained through the second parametrization. It follows that given a surface, we don’t need to stick to any unique parametrization, but, when integrating vector fields, we need to decide in advance in which direction the normal will point and Then choose any parameter corresponding to that direction.

Another issue is that sometimes surfaces do not have parametrizations that cover the entire surface. The obvious solution is then to split that surface into multiple pieces, calculate the integral of the surface on each piece, and then add them all up. This is exactly how things work, but when integrating vector fields, one again has to 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 result is be compatible. For the cylinder, this means that if we decide that the normal for the side sphere will protrude from the body, then for the top and bottom circular parts, the normal must also exit the body.

Finally, there are surfaces that do not consider a surface normal at every point with consistent results (for example, the Möbius strip). If such a surface is divided into pieces, a parametrization on each piece and the corresponding surface normal is chosen, and the pieces are put back together, we will find that the normal vectors coming from the different pieces match cannot be done. This means that at any junction between two pieces we will have normal vectors pointing in opposite directions. Such a surface is said to be non-orientable, and on such a surface, one cannot speak of integrating vector fields.