Ostrogradsky formula for surface integrals. Gaussian formula

Let's consider the field of a point charge $q$ and find the flow of the intensity vector ($\overrightarrow(E)$) through the closed surface $S$. We will assume that the charge is located inside the surface. The flux of the tension vector through any surface is equal to the number of lines of the tension vector that go out (start at the charge, if $q>0$) or the number of lines $\overrightarrow(E)$ going in, if $q \[Ф_E=\frac( q)((\varepsilon )_0)\ \left(1\right),\]

where the sign of the flux coincides with the sign of the charge.

Ostrogradsky-Gauss theorem in integral form

Let us assume that inside the surface S there are N point charges, values $q_1,q_2,\dots q_N.$ From the principle of superposition we know that the resulting field strength of all N charges can be found as the sum of the field strengths that are created by each of the charges, then There is:

Therefore, for the flow of a system of point charges we can write:

Using formula (1), we obtain that:

\[Ф_E=\oint\limits_S(\overrightarrow(E)d\overrightarrow(S))=\frac(1)((\varepsilon )_0)\sum\limits^N_(i=1)(q_i\ )\ left(4\right).\]

Equation (4) means that the flow of the electric field strength vector through a closed surface is equal to the algebraic sum of the charges that are inside this surface, divided by the electric constant. This is the Ostrogradsky-Gauss theorem in integral form. This theorem is a consequence of Coulomb's law. The significance of this theorem is that it allows one to quite simply calculate electric fields for various charge distributions.

As a consequence of the Ostrogradsky-Gauss theorem, it must be said that the flux of the intensity vector ($Ф_E$) through a closed surface in the case in which the charges are outside this surface is equal to zero.

In the case where the discreteness of charges can be ignored, the concept of volumetric charge density ($\rho $) is used if the charge is distributed throughout the volume. It is defined as:

\[\rho =\frac(dq)(dV)\left(5\right),\]

where $dq$ is a charge that can be considered point-like, $dV$ is a small volume. (Regarding $dV$, the following remark must be made. This volume is small enough that the charge density in it can be considered constant, but large enough so that charge discreteness does not begin to appear). The total charge that is in the cavity can be found as:

\[\sum\limits^N_(i=1)(q_i\ )=\int\limits_V(\rho dV)\left(6\right).\]

In this case, we rewrite formula (4) in the form:

\[\oint\limits_S(\overrightarrow(E)d\overrightarrow(S))=\frac(1)((\varepsilon )_0)\int\limits_V(\rho dV)\left(7\right).\ ]

Ostrogradsky-Gauss theorem in differential form

Using the Ostrogradsky-Gauss formula for any field of vector nature, with the help of which the transition from integration over a closed surface to integration over a volume is carried out:

\[\oint\limits_S(\overrightarrow(a)\overrightarrow(dS)=\int\nolimits_V(div))\overrightarrow(a)dV\ \left(8\right),\]

where $\overrightarrow(a)-$field vector (in our case it is $\overrightarrow(E)$), $div\overrightarrow(a)=\overrightarrow(\nabla )\overrightarrow(a)=\frac(\partial a_x)(\partial x)+\frac(\partial a_y)(\partial y)+\frac(\partial a_z)(\partial z)$ -- divergence of the vector $\overrightarrow(a)$ at the point with coordinates ( x,y,z), which maps a vector field to a scalar one. $\overrightarrow(\nabla )=\frac(\partial )(\partial x)\overrightarrow(i)+\frac(\partial )(\partial y)\overrightarrow(j)+\frac(\partial )(\ partial z)\overrightarrow(k)$ - observable operator. (In our case it will be $div\overrightarrow(E)=\overrightarrow(\nabla )\overrightarrow(E)=\frac(\partial E_x)(\partial x)+\frac(\partial E_y)(\partial y) +\frac(\partial E_z)(\partial z)$) -- divergence of the tension vector. Following the above, we rewrite formula (6) as:

\[\oint\limits_S(\overrightarrow(E)\overrightarrow(dS)=\int\nolimits_V(div))\overrightarrow(E)dV=\frac(1)((\varepsilon )_0)\int\limits_V( \rho dV)\left(9\right).\]

The equalities in equation (9) are satisfied for any volume, and this is only feasible if the functions that are in the integrands are equal in each current of space, that is, we can write that:

Expression (10) is the Ostrogradsky-Gauss theorem in differential form. Its interpretation is as follows: charges are sources of an electric field. If $div\overrightarrow(E)>0$, then at these points of the field (charges are positive) we have field sources, if $div\overrightarrow(E)

Assignment: The charge is uniformly distributed over the volume; a cubic surface with side b is selected in this volume. It is inscribed in the sphere. Find the ratio of the tension vector fluxes through these surfaces.

According to Gauss's theorem, the flux ($Ф_E$) of the intensity vector $\overrightarrow(E)$ through a closed surface with a uniform charge distribution over the volume is equal to:

\[Ф_E=\frac(1)((\varepsilon )_0)Q=\frac(1)((\varepsilon )_0)\int\limits_V(\rho dV=\frac(\rho )((\varepsilon ) _0)\int\limits_V(dV)=\frac(\rho V)((\varepsilon )_0))\left(1.1\right).\]

Therefore, we need to determine the volumes of the cube and the ball if the ball is described around this cube. To begin with, the volume of a cube ($V_k$) if its side b is equal to:

Let's find the volume of the ball ($V_(sh)$) using the formula:

where $D$ is the diameter of the ball and (since the ball is circumscribed around the cube), the main diagonal of the cube. Therefore, we need to express the diagonal of a cube in terms of its side. This is easy to do if you use the Pythagorean theorem. To calculate the diagonal of a cube, for example, (1.5), we first need to find the diagonal of the square (the lower base of the cube) (1.6). The length of the diagonal (1.6) is equal to:

In this case, the length of the diagonal (1.5) is equal to:

\[(D=D)_(15)=\sqrt(b^2+((\sqrt(b^2+b^2\ \ \ )))^2)=b\sqrt(3)\ \left (1.5\right).\]

Substituting the found diameter of the ball into (1.3), we obtain:

Now we can find the fluxes of the tension vector through the surface of the cube, it is equal to:

\[Ф_(Ek)=\frac(\rho V_k)((\varepsilon )_0)=\frac(\rho b^3)((\varepsilon )_0)\left(1.7\right),\]

through the surface of the ball:

\[Ф_(Esh)=\frac(\rho V_(sh))((\varepsilon )_0)=\frac(\rho )((\varepsilon )_0)\frac(\sqrt(3))(2) \pi b^3\ \left(1.8\right).\]

Let's find the ratio $\frac(Ф_(Esh))(Ф_(Ek))$:

\[\frac(Ф_(Esh))(Ф_(Ek))=\frac(\frac(с)(\varepsilon_0)\frac(\sqrt(3))(2) \pi b^3)(\frac (сb^3)(\varepsilon_0))=\frac(\pi)(2)\sqrt(3)\ \approx 2.7\left(1.9\right).\]

Answer: The flux through the surface of the ball is 2.7 times greater.

Task: Prove that the charge of a conductor is located on its surface.

We use Gauss's theorem to prove it. Let us select a closed surface of arbitrary shape in the conductor near the surface of the conductor (Fig. 2).

Let us assume that there are charges inside the conductor, we write the Ostrogradsky-Gauss theorem for field divergence for any point on the surface S:

where $\rho is the density\ $of the internal charge. However, there is no field inside the conductor, that is, $\overrightarrow(E)=0$, therefore, $div\overrightarrow(E)=0\to \rho =0$. The Ostrogradsky-Gauss theorem in differential form is local, that is, it is written for a field point, we did not select the point in a special way, therefore, the charge density is zero at any point in the field inside the conductor.

Gauss-Ostrogradsky formula.

Let us define in space a closed three-dimensional region V, bounded by the surface S and projected onto the Oxy plane into the regular region D.

Let us define continuous functions P(x, y, z), Q(x, y, z) and R(x, y, z) at each point of the region V and surface S and calculate the integral

Let us set the orientation of the surface S by choosing the direction of the external normal, then on S1

cos(n, z)< 0, на S2 cos(n, z) >0, a on S3 cos(n, z) = 0. The double integrals on the right side of the previous equality are equal to the corresponding surface integrals:

(The “-” sign in the second integral appears due to the fact that the elements of surface area S1 and area D are related by the relation dxdy = ΔS(-cos(n, z))). Therefore, the original integral can be represented as:

The final result can be written like this:

In the same way we can obtain the relations

Adding these three equalities, we obtain the Gauss-Ostrogradsky formula:

Using formula 13.9, which defines the connection between surface integrals of the 1st and 2nd kind, we can write the Gauss-Ostrogradsky formula in a different form:

where the notation “S+” means that the integral on the right is calculated over the outside of the surface S.

Divergence of a vector field.

Let's continue studying the characteristics of vector fields.

Definition 15.1. The divergence of a vector field A = (Ax, Ay, Az), where Ax, Ay, Az are functions of x, y, z, is called

. (15.3)

Remark 1. From the definition it is clear that divergence is a scalar function.

Remark 2. The word “divergence” means “divergence”, since divergence characterizes the density of sources of a given vector field at the point in question.

Let us consider the Gauss-Ostrogradsky formula taking into account the definitions of flow and divergence of a vector field. Then on the left side of formula (15.1) there is a triple integral over the volume V of the divergence of the vector field (P, Q, R), and on the right side there is the flow of this vector through the surface S bounding the body:

(15.4)

Let us prove that the magnitude of divergence at a given point does not depend on the choice of coordinate system. Let's consider a certain point M, which is surrounded by a three-dimensional region V, limited by the surface S. Let's divide both sides of formula (15.4) by V and go to the limit when the body V is contracted to the point M. We obtain:

. (15.5)

Stokes formula.

Let us consider a surface S such that any straight line parallel to the Oz axis intersects it at one point. Let us denote the boundary of the surface λ and choose as the positive direction of the normal one in which it forms an acute angle with the positive direction of the Oz axis. If the surface equation has the form z = f(x, y), then the direction cosines of the normal are given by the formulas

, ,

Let us consider some three-dimensional region V, in which the entire surface S lies, and define in this region a function P(x, y, z), continuous along with first-order partial derivatives. Let us calculate the curvilinear integral of the 2nd kind over the curve λ:

The equation of line λ has the form z = f(x, y), where x, y are the coordinates of the points of line L, which is the projection of λ onto the Oxy plane (Fig. 2). Therefore, using formula (10.8), we obtain:

Let us denote P(x, y) = P(x, y, f(x, y)), Q(x, y) = 0 and apply Green’s formula to the integral on the right side of the previous equality:

where the region D is limited by the line L. Let us transform the left integrand using the formula for the derivative of a complex function:

and substitute it into the previous equality:

. Then

=Now we apply formula (13.7) to the integrals on the right and move on to surface integrals of the 1st kind over the surface σ:

because . Therefore, the final result of the transformation looks like this:

In this case, the direction of traversing the contour λ is selected corresponding to the positive direction of the normal (Fig. 2).

By defining continuously differentiable functions Q(x, y, z) and R(x, y, z) in the domain V, we can obtain similar relations for them:

Adding the left and right sides of the resulting equalities, we obtain the Stokes formula, which establishes a connection between the surface integral of the 1st kind over the surface σ and the curvilinear integral of the 2nd kind along the contour λ bounding it, taking into account the orientation of the surface:

(15.6)

The last entry allows you to better remember the integrand on the right side of the Stokes formula, which can be obtained by expanding the determinant in the first line and taking into account that in its second line there are operators of partial differentiation with respect to the corresponding variables, applied to the functions in the third line.

Using the connection between surface integrals of the 1st and 2nd kind (formula (13.9)), we can write the Stokes formula in a different form:

Vector field rotor.

Definition 15.2. The rotor or vortex vector of a vector field A = (Ax, Ay, Az), where Ax, Ay, Az are functions of x, y, z, is a vector defined as follows:

. (15.8)

Remark 1. The rotor characterizes the vorticity of the field A at a given point, that is, the presence of rotational movements, since its module is equal to twice the angular velocity at this point.

Remark 2. The Stokes formula in the vector formulation has the form:

, (15.9)

that is, the circulation of a vector along a closed contour is equal to the rotor flux of this vector through a surface stretched over a given contour.

Remark 3. Another, invariant, definition of a rotor can be given. To do this, consider an arbitrary direction n emanating from a given point M, and surround this point with a flat area σ perpendicular to n and limited by the contour λ. Applying the Stokes formula, we obtain:

Dividing both sides of this equality by σ and contracting the area σ to a given point, we find in the limit that

Thus, it is possible to determine the projection of the rotor onto any axis, that is, the vector rot A does not depend on the choice of coordinate system.

And therefore, if you came from a search engine, then please start from the first part, where we examined in detail and solved an important problem. Namely, we found the flow of a vector field through a closed surface in the direction of its outer normal:

In the course of a long, long solution, we received the answer, which, within the framework of a conditional hydrodynamic model, means the following: how much liquid is in unit of time entered the pyramid - so much flowed out of it.

However, this is not always the case, and in practice the flow often turns out to be positive or negative. Let's think about the meaningful meaning of these results and, for greater clarity, consider not a pyramid, but a piece of a river, limited externally oriented surface and velocity field of this river in the region.

Let us assume that the flow through the closed surface turns out to be positive: . What does this mean? It means that per unit of time MORE fluid leaked out of the area than went in. Therefore, somewhere in the area there is source(s) fields. This could be, for example, a river inflow that increases its speed, or simply someone poured a bucket of water.

A negative flux through a closed surface tells us that per unit of time the area has “absorbed” liquid (more came in than came out). And the reason for this is drain(s) fields in this area. For example, an underground cave or a pump that pumps out water.

And finally, with zero flow, two situations are possible: either in the area No sources And drains, or they compensate each other.

By the way, mutual compensation most often occurs in the first two cases. So, for example, if , then this does not mean that there are no drains. Perhaps the sources turned out to be more powerful, and as a result per unit of time 5 units of liquid splash out through the surface.

And therefore, there is an interest in finding out whether the vector field has sources/sinks, and if so, where. And a tricky science called mathematical analysis will help us with this.

Consider some point in the region and its infinitesimal closed neighborhood (such as a sphere or cube). Vector field flow through the surface this neighborhood in the external direction is called divergence fields in at this point, and is denoted by . And here there is no escape from exposure:

– if , then the vector field has source at this point (its infinitesimal neighborhood);

– if , then drain;

– and if , then there are no sources or sinks at the point.

Further. How to find this very divergence? If in every point region, a vector field is defined and its components are differentiable at these points, then scalar function divergence has the following form:

or, as it is written more briefly:

Thus, in the area vector field is put in accordance scalar field its divergence.

And here we can immediately highlight a special case. A field whose divergence is zero AT ALL points in the region is called non-divergent or solenoidal. This means that it has no sources or sinks. An example is often given of a donut pipe with circulating water that does not disappear anywhere, and new water does not appear there. But an even more illustrative example is the magnetic field with its closed power lines, which have no beginning and end.

Fine. The function allows us to calculate the divergence at individual points, and the question arises: is it possible to calculate the total divergence over the entire body?

...did you ever think you'd be so excited about triple integrals? =)

Let's get back to the epic Example 1, where we have a zero flow through the pyramid, and calculate the divergence of the vector field. It is obvious that the field itself and derivatives its components are defined not only in the pyramid, but in general throughout space:

Let's create a scalar divergence function, or as they often say, find the divergence:

The resulting function assigns a zero to each point in space, which means the vector field is everywhere solenoidal. According to the Gauss-Ostrogradsky formula, the vector field flux through the outer side of the pyramid is equal to:

Note : because the field is divergence-free throughout space, then the flux is zero through any closed surface

However, you shouldn’t be upset, because if you were required to calculate the flow using the first method, then there’s no escape =) And, by the way, they often do.

And here we also need to emphasize the following: if you calculated the flow through a closed surface, and you got zero, then this doesn't mean yet that there are no sources or drains in the area. They may exist, but they compensate each other. And the first solution does not give us an answer to this question.

Therefore, we solve the second example using the second method :)

Example 2

Check whether the vector field is solenoidal and find its flow through a closed surface using the Gauss-Ostrogradsky formula

The results should match. Please note that checking the field for solenoidality is an integral part of the assignment, and this question must be given a reasoned written answer. An approximate sample solution is at the end of the lesson, and what’s nice is that the problem can be written in a minimalist style, without unnecessary notation and even without writing down the formula itself.

Well, now I’ll tell you, or rather remind you, of the universal method for finding normal surface vectors:

Example 3

Given a vector field and a closed surface. Calculate the vector field flux through a given surface in the direction of the outer normal:

a) directly;
b) according to the Gauss-Ostrogradsky formula.

A common formulation that allows you to once again realize the full value of the formula =)

Solution: the drawing here is simple:

but here’s the solution – “pipe” =)

a) Let us find the flux of the vector field through the full surface of the cylinder in the direction of the outer normal directly. By virtue of additivity surface integral:

- side surface cylinder ;
– its lower base (unit circle in the plane);
– and the upper base (unit circle in the plane).

1) A cylindrical surface is parallel to the axis and the question arises, how to find its normal vectors? Very simple. Vector normal to surface at a point is given as follows:

In this case:

Thus, we obtain an entire function of normal vectors for various points of the cylinder:

But we need unit vectors. They are wanted as standard:

Control:

Yes, let's make sure that they are “looking” outward. To do this, you can take several specific points on the surface (easiest in the plane) and see what vectors you get. So, for example, for a point we get:
- all OK. Actually, this vector is shown in the drawing as an example. Check some other points yourself and make sure that you get vectors in the desired direction.

and reduce the solution to a surface integral of the 1st kind:

In this case, the plane is not suitable for projection. Why? Because a cylindrical surface will be projected into circle zero area and you get zero. But the field vectors stick out from the side surface, and a flow can easily flow through it!

Therefore, we have two coordinate planes at our disposal; I will choose the more visual frontal plane for projection. And here another difficulty arises - a cylindrical surface, which means that the resulting integral of the 1st kind will have to be divided into 2 parts:
, Where:

- the piece of cylinder closest to us, and - the farthest piece of it.

Let's carry out calculations for the first integral:

We use the appropriate formula:
, Where:

According to the formula:

The projection onto the plane is obvious:

Let us choose the following order of traversal of the area:

When calculating the second integral, you get exactly the same result:

Thus:

I gave a long general solution (just in case), but in fact there is a short and elegant way - you can immediately substitute And :

and, according to geometric meaning of these integrals, this sum is equal to the area of the lateral surface of the cylinder:

Knowledge is power =)

2) Let us calculate the flux of the vector field through oriented unit circle.

With the normal and scalar product everything is simple:

and with the surface integral it’s even simpler:
, because the

3) The third integral begins similarly:

We use the formula, in this case:

Projection (surfaces to plane) represents a circle of area, and according to geometric meaning of the integral :

And finally, flow across the entire surface:

Answer:

What does this result mean, by the way? Positive flux through the outer surface means that there are field sources inside the cylinder. Otherwise, where would the units of liquid that leaked out come from? (per unit of time)

b) Let’s solve the problem using the Gauss-Ostrogradsky formula:

And, first of all, here you need to make sure that the components and their derivatives determined in all points of the body. Otherwise, the formula cannot be applied! I must warn you that this is not an empty formality - in practice, there are fields with roots and logarithms, and this is where problems can arise.

Let's create a divergence function:
, which is very useful to analyze:

As “z” increases from 0 to 2, the divergence is strictly positive and increases. This means that, firstly, inside the cylinder there are exclusively field sources. And, secondly, these sources are amplified, i.e. the fluid flowing from bottom to top begins to accelerate. Therefore, we can immediately say that the flow through the outer surface will be positive. What we will now once again verify analytically:

Since the projection of a body onto a plane is a circle of unit radius (I won’t draw it), it is convenient to go to cylindrical coordinate system:

Thus, using “rho”, “theta” and “phi” one can uniquely determine any point in space.

Where is the spherical coordinate system used? Well, of course, in astronomy. But it also found its modest application in

formulas related to various branches of mathematics and bearing the name of K. Gauss.

1) Quadrature G. f. - formulas of the form

in which the nodes x k and odds A k independent of function f(x) and are chosen so that the formula is accurate (i.e. Rn= 0) for an arbitrary polynomial of degree 2n - 1. Unlike the Newton-Cotes quadrature formulas, the nodes in quadrature geometric functions, generally speaking, are not equidistant. If p(x) ≥ 0 And

then for any natural n there is a single quadrature geometric function. These formulas are of great practical importance, because in some cases they provide significantly greater accuracy than quadrature formulas with the same number of equally spaced nodes. Gauss himself investigated (1816) the case p(x) ≡ 1.

2) G. f., expressing total curvature (See Total curvature) To the surface through the coefficients of its linear element; in coordinates for which ds 2 = λ(du 2 + dv 2), G. f. looks like

This formula was published in 1827 and shows that the total curvature does not change when the surface is bent. It constitutes the content of one of the main proposals of the internal geometry created by Gauss (See Internal geometry) of the surface.

3) G. f. for Gaussian sums:

This formula was used by Gauss (1801) in one of the proofs of the law of reciprocity of quadratic residues (See Quadratic residue)

Where R And q- odd prime numbers, and

4) G. f. for the sum of a hypergeometric series (See Hypergeometric series). If Re (c - b - a) > 0, That

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"Gauss formula" in books

4. Formulas

From the book The Most Self author Losev Alexey Fedorovich

4. Formulas The dialectic of the first symbol of the symbol of being, or the existential symbol, ends with the category of emanation. All that remains now is to summarize it in the most concise theses that do not contain anything superfluous.I.1. a) Being is being. If being is only being and more

Chapter 5 APPLICATION OF GAUSS'S LAW

From book 5a. Electricity and magnetism author Feynman Richard Phillips

Appendix Easter tables and tables of dates of the first spring astronomical full moons, calculated using Gauss formulas (G.V. Nosovsky)

From the book Easter [Calendar-astronomical investigation of chronology. Hildebrand and Crescentius. Gothic War] author Nosovsky Gleb Vladimirovich

Appendix Easter tables and tables of dates of the first spring astronomical full moons, calculated using Gaussian formulas (G.V. Nosovsky) An asterisk (*) in the last column marks the years when the calendar Orthodox Easter determined by Paschal would have been celebrated earlier