applications of divergence (mini project)

8/11/2019 Applications of divergence (mini project)

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Submitted by,

V KARTHIK REDDY 12BEC0440

V BHARGAV CHOWDARY 12BEI0015

SHIVENDRE PRATAP SINGH 12BEC0230

SHIANGH SINGH PUNDHIR 12BEC0334

Mini ProjectApplications of Divergence

This project is about Divergence and its applications in

various fields.
http://en./wikipedia/applications+of+divergence


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APPLICATIONS OF DIVERGENCE

Divergence theorem was first discovered

by Lagrange in 1762, then later independentlyrediscovered by Gauss in 1813, by Green in

1828, and in 1826 by Ostrogradsky, who also

gave the first proof of the theorem.

Subsequently, variations on the divergence

theorem are correctly called Ostrogradsky's

theorem, but also commonly Gauss's theorem,

or Green's theorem.


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INTRODUCTION :

The divergence theorem, more commonly known especially in older literature as Gauss'stheorem (e.g., Arfken 1985) and also known as the Gauss-Ostrogradsky theorem, is a

theorem in vector calculus that can be stated as follows. Let be a region in space with

boundary . Then the volumeintegral of the divergence of over and the surface

integral of over the boundary of are related by

The divergence theorem is a mathematical statement of the physical fact that, in the

absence of the creation or destruction of matter, the density within a region of space

can change only by having it flow into or away from the region through its boundary.

A special case of the divergence theorem follows by specializing to the plane. Letting

be a region in the plane with boundary , above equation then collapses to

If the vectorfield satisfies certain constraints, simplified forms can be used. For

example, if where is a constant vector , then

But

so

and


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But , and must vary with so that cannot always equal zero. Therefore,

Similarly, if , where is a constant vector , then

Required to Prove: The Divergence theorem. If V is the volume bounded by aclosed surface S and A is a vector function of position with continuous derivatives, then

where n is the positive (outward drawn) normal to S.


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Proof: The Divergence theorem in the full generality in which it is stated is not easy toprove. However given a sufficiently simple region it is quite easily proved.

Let S be a closed surface so shaped that any line parallel to any coordinate axis cuts the

surface in at most two points. We will now proceed to prove the following assertion:

Denote the projection of the surface on the xy plane by R. A line erected from within R

perpendicular to the xy plane intersects the surface at two points, at a lower surface and

an upper surface. Denote the lower surface by S1and the upper surface by S2. Let the

equation of S1be z = f1(x, y) and the equation of S2be z = f2(x, y). See Fig. 1. Now

Now by the Fundamental Theorem of Integral Calculus

Substituting 3) into the right member of 2) we get


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For the upper surface S2, dy dx = cos2dS2= kn2dS2since the normal n2to S2makesan acute angle 2with k.

For the lower surface S1, dy dx = -cos1dS1= -kn1dS1since the normal n1to S1makesanobtuse angle1with k.

Consequently

Substituting 5) and 6) into 4) we get


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or

which is what we wished to prove.

In the same way, by projecting S on the other coordinate planes, we can obtain

Adding 8), 9) and 10) we get

or

SOME EXAMPLES ON DIVERGENCE THEOREM:

Example 1:

ComputeSFdSwhere


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F = (3x+z77, y2sinx2z, xz+yex5)andSis surface of box

0x1,0y3,0z2.Use outward normal n.

For F = (xy2,yz2,x2z), use the divergence theorem to evaluate

SFdS

solution:Given the ugly nature of the vector field, it would be hard to compute thisintegral directly. However, the divergence ofFis nice:

divF = 3+2y+x.

We use the divergence theorem to convert the surface integral into a triple integral

SFDs =BdivFdVWhere Bis the box

0x1,0y3,0z2.

We compute the triple integral of divF = 3+2y+xover the box B:

SFdS=103020(3+2y+x)dzdydx=1030(6+4y+2x)dydx=10(18+18+6x)dx=36+3=39

.

Example 2: Use the divergence theorem to calculate , where Sis the surface of the

box Bwith vertices (1, 2, 3) with outwards pointing normal vector and F(x, y, z) = (x2z

3,

2xyz3, xz

4).

Solution: Note that the surface integral will be difficult to compute, since there are six different

components to parameterize (corresponding to the six sides of the box) and so one would have

to compute six different integrals. Instead, using Gauss Theorem, it is easier to compute the

integral (F)of B.

First, we compute (F) = 2xz3+ 2xz

3+ 4xz

3= 8xz

3. Now we integrate this function over the

region B bounded by S


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APPLICATIONS OF DIVERGENCE THEOREM:

Divergence theorem has a wide applications on differential form and integral

form of physical law and continuity equations.

Divergence theorem has a wide applications on inverse square laws.

Divergence theorem of gauss plays a central role in deviation of differential

equation in fluid dynamics, electrodynamics, gravitational fields and optics.

There are some physical applications like fluid flow and heat transfer.

Divergence theorem is also used in Archimedes principle and Laplace equation.

APPLICATION OF DIVERGENCE THEOREM IN ARCHIMEDES PRINCIPLE:

Gauss divergence theorem may be used to derive Archimedes principle for the buoyant

force on a body totally immersed in a fluid of constant density (independent of depth).


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Examine an elementary section of the surface Sof the immersed body, at a depth z< 0

below the surface of the fluid:

The pressure at any depth z is the weight of fluid per unit area from the column of fluid

above that area. Therefore

pressure = p = g z g is the weight of the column

zis the height of the column (note z< 0).

The normal vectorN to Sis directed outward, but the hydrostatic force on the surface

(due to the pressure p) acts inward. The element of hydrostatic force on S is

pressure area direction g z S g z S N N

The element of buoyant force on S is the component of the hydrostatic force in the

direction of k(vertically upwards):


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g z S N kg

Define

g zF k and

dSdS N .

Summing over all such elements S, the total buoyant force on the immersed object is

S S V

g z dS dV k N F dS Fvv v v

g g g (by the Gauss Divergence Theorem)

, , 0, 0,V V

g z dV g z dV

x y z

k

vg g

provided 0V

g dV gz

=weight of the fluid displaced.

Therefore the total buoyant force on an object fully immersed in a fluid equals the weight of

the fluid displaced by the immersed object (Archimedes principle).

APPLICATION OF DIVERGENCE THEOREM : LAPLACE EQUATION

The combination div grad, ( ) or is called the "Laplacian" differential

operator,

The equation ( ) f = 0 is called Laplace's equation. Static electric and steady state magnetic

fields obey this equation where there are no charges or current.

Any solution to this equation in R has the property that its value at the center of a sphere within

R is the average of its value on the sphere's surface. If g(r) obeys Laplace's equation inside a

spherical surface, S, of radius a, centered at r', we have


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Laplace's Equation

Thus solutions to Laplace's equation are very smooth they have no bumps maxima or

minima in R and essentially "interpolate" smoothly between their values on the

boundaries of R. We prove this important fact as an application of the divergence

theorem.

This result also implies that if we know the divergence of a vector v and its curl

everywhere, these are differentiable everywhere, and vvanishes at infinity, then vis

uniquely determined. proof box( if there were two solutions v andv'with the same

divergence and curl, then on applying the double cross product identity we find thatevery component of their difference obeys Laplace's equation everywhere. Its value

anywhere is then its average value on a huge circle at infinity, which is 0 by assumption.

The same conclusion holds if vand v'are required to behave at infinity in the same way,

so that v -v'must approach 0 for large arguments.)

PROOF:

Suppose our function, f(r), obeys Laplace's equation within some sphere S centered

at r':

div gradf = ( ) f = 0 inside S

We apply the divergence theorem to the vectorf g - g f in the sphere with surface S

excluding a tiny sphere of radius b with surface S' having the same center. We obtain

the latter being obtained by substituting for g. The second integral on the second line

vanishes here as can be seen by applying the divergence theorem again within S and

noticing that the Laplacian applied to f is 0.

The right hand side here is the average value of f on S. The similar integral over S' is

evaluated in exactly the same way and is the average value of f on S'. We conclude that

the average value of v on any sphere with center at r' is the same. Obviously as the


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sphere approaches radius 0that average value becomes the value of f at r'.

The method used in this argument is a very important and general one that is used in

dealing with many differential equations. In fact the use of the divergence theorem in

the form used above is often called "Green's Theorem." And the function g defined

above is called a "Green's function" for Laplaces's equation.We can use this function g to find a vector field vthat vanishes at infinity obeying

div v= , curlv = 0. (we assume that r is sufficiently well behaved, integrable, vanishes

at infinity etc...) Suppose we write vas grad f.

We get

(This equation is called Poisson's equationand is obeyed by the potential produced by a

charge distribution with charge density .)

A solution that vanishes at infinity is then given by

PROOF :

The same approach can be used to obtain a vector potential A and determine a

vector vobeying

v = 0, v =j

Withv = A we obtain

( ( A)) = j

and with the double cross identity

( ( A))= ( A)-( ) A

and the "gauge" condition (which we are free to assume) ) A = 0, we find that each

component of Aobeys Poisson's equation with source - the corresponding component of j.

We may therefore find a formula for each coordinate of Aexactly like the corresponding

formula for the scalar potential V. Again vmay be recovered from Aby differentiating.

These results are of some use in the study of electromagnetic fields but they don't solve all


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problems. Often the known charges and currents induce unknown charges and currents in

conducting surfaces, and one wants to determine the fields, in circumstances in which there are

some unknown charges and /or currents and you know conditions on the field at the conductor

surfaces instead of the charges and currents in them.

In this way laplace equation can be verified using divergence theorem.

AERODYNAMIC APPLICATION OF DIVERGENCE THEOREM:

The Aerodynamic Continuity Equation

1. The surface integral of mass flux around a control volume without sources or sinks is

equal to the rate of mass storage.

2. If the flow at a particular point is incompressible, then the net velocity flux around the

control volume must be zero.

3.

As net velocity flux at a point requires taking the limit of an integral, one instead merelycalculates the divergence.

4. If the divergence at that point is zero, then it is incompressible. If it is positive, the fluid

is expanding, and vice versa.

Gausss Theorem can be applied to any vector field which obeys an inverse-square law

(except at the origin) such as gravity, electrostatic attraction, and even examples in

quantum physics such as probability density.

PRACTICAL APPLICATIONS OF DIVERGENCE THEOREM:

Remember that the divergence of a field f at a point p measures the tendency of the

underlying substance, or energy, to expand at the point p. If f is the velocity of a fluid, is the

fluid expanding (positive divergence), just passing through (0 divergence), or compressing

(negative divergence)? Note that f could cause water to move slower, but over a larger area,

and still give a divergence of 0. Let water rush into the neck of a funnel at high speed, and

move out the wide mouth at a slower speed. A depiction of f might present long arrows

clustered together, pushing the water quickly into the neck of the funnel, and short arrows

spread apart around the mouth of the funnel. It is possible for f to have divergence 0

throughout. At any point p, water comes in quickly from behind, and leaves at a slower speed

and in several directions. Water does not compress or expand at the point p; it is merelypassing through. In this sense, referring to divergence as "spreading out" is somewhat

misleading. The water does indeed spread out in space, but it does not actually expand. It

moves slower as it spreads out, to compensate. The same amount of material is involved from

start to finish.


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Here is another 0 divergence example. Consider the electric field, emanating from an

electron at the origin. This is an inverse square field. This can be written as some constant

(which we will ignore) times x y or z over r3, where r is the radial distance

sqrt(x2+y2+z2). The partial with respect to x is 1/r3 - 3x2/r5. Do the same for y and z and

add them up, giving 3/r3 - 3r2/r5, or 0. The electric field gets weaker as you move awayfrom its source, but it gets weaker in proportion to the surface area of the expanding

shell. This is the inverse square law, and it implies a divergence of 0 everywhere, except at

the origin, where 1/r is not defined. Electric charge bursts into existence at the origin, and

charge is conserved everywhere else.

These are some real life examples related to divergence theorem.

Differential and Integral Operators use in Divergence Theorem:

One of the most important and useful mathematical constructs is the "del operator",

usually denoted by the symbol (which is called the "nabla"). This can be regarded as avector whose components in the three principle directions of a Cartesian coordinate

system are partial differentiations with respect to those three directions. Of course, the

partial differentiations by themselves have no definite magnitude until we apply them tosome function of the coordinates. Letting i,j, kdenote the basis vectors in the x,y,z

directions, the del operator can be expressed as

All the main operations of vector calculus, namely, the divergence, the gradient, the curl,

and the Laplacian can be constructed from this single operator. The entities on which we

operate may be either scalar fields or vector fields. A scalar field is just a single-valuedfunction of the coordinates x,y,z. For example, the static pressure of air in a certain

region could be expressed as a scalar field p(x,y,z), because there is just a single value of

static pressure p at each point. On the other hand, a vector field assigns a vector vto eachpoint in space. An example of this would be the velocity v(x,y,z) of the air throughout a

certain region.

If we simply multiply a scalar field such as p(x,y,z) by the del operator, the result is a

vector field, and the components of the vector at each point are just the partial derivatives

of the scalar field at that point, i.e.,

This is called the gradient of p. On the other hand, if we multiply a vector field v(x,y,z)

by the del operator we first need to decide what kind of "multiplication" we want to use,

because there are two different kinds of vector multiplication, commonly called the dot


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product and the cross product. For two arbitrary vectors a= axi+ ayj+ azk and b= bzi

+ byj+ bzk the dot product aband the cross product abare defined as

Intuitively, the dot product is a scalar equal to the product of the magnitudes of aand b

times the cosine of the angle between them, and the cross product is a vector

perpendicular to both aand b(with direction determined conventionally by the "righthand rule") and whose magnitude is equal to the product of the magnitudes of aand b

times the sine of the angle between them. Since the dot product yields a scalar, it is often

called the "scalar product". Likewise the cross product is often called the "vector

product".

The dot product of and a vector field v(x,y,z) = vx(x,y,z)i + vy(x,y,z)j + vz(x,y,z)kgives a scalar, known as the divergence of v, for each point in space:

The cross product of and a vector field v(x,y,z) gives a vector, known as the curl of v,

for each point in space:

Notice that the gradient of a scalar field is a vector field, the divergence of a vector fieldis a scalar field, and the curl of a vector field is a vector field. Can we construct from the

del operator a natural differential operator that creates a scalar field from a scalar

field? Actually we already have the ingredients for such an operator, because if we applythe gradient operator to a scalar field to give a vector field, and then apply the divergence

operator to this result, we get a scalar field. This is sometimes called the "div grad" of a

scalar field, and is given by

For convenience we usually denote this operator by the symbol 2, and it is usually

called the Laplacian operator, because Laplace studied physical applications of scalar

fields (x,y,z) (such as the potential of an inverse-square force law) that satisfy the

equation , i.e.,


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This formula arises not only in the context of potential fields (such as the electro-staticpotential in an electric field, and the velocity potential in a frictionless ideal fluid), it also

appears in Poisson's equation and in the fundamental wave equation

where c is the speed of propagation of the wave. Incidentally, this suggests another

useful differential operator, formed by bringing the right hand term over to the left side togive

This operator is known as the d'Alembertian (because the wave equation was first studiedby Jean d'Alembert), and it could be regarded as the dot product of two "dal" operators

where i,j,k,lare basis vectors of a four-dimensional Cartesian coordinate system.

One of the most important theorems in vector analysis is known as the Divergence

Theorem, which is also sometimes called Gauss' Theorem. This is essentially just anapplication of the fundamental theorem of calculus

This enables us to express the integral of the quantity df/dx along an interval in terms ofthe values of f itself at the endpoints of that interval. Now suppose we are given a scalar

function f(x,y,z) throughout a region V enclosed by a surface S, and we want to evaluate

the integral of the quantity f/x over this entire region. This can be written as


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where the three integrals are evaluated over suitable ranges to cover the entire region

V. For any fixed values of y = y0and z = z0the function f(x,y0,z0) is totally dependent on

x, so we can evaluate the integral along any line parallel to the x axis through the regionV for any particular y0,z0using the fundamental theorem of calculus

where a,b are the x values at which the line intersects the surface S. (For ease ofdescription we are assuming every line parallel to the x axis intersects the surface in only

two points, but this restriction turns out to be unnecessary.) We now just need to

integrate the above quantity over suitable ranges of y0and z0to give the integral

throughout the entire region V. Notice that the two points (a,y0,x0) and (b,y0,x0) are both,by definition, on the surface S, at opposite ends of a line interval parallel to the x

axis. Hence, we can cover all the contributions by integrating the value of the function

f(x,y,z) over the entire surface S.

Of course, we recognize that f(a,y0,z0) is subtracted from the total, whereas f(b,y0,z0) is

added, so it's clear that we need to apply a weighting factor to f(x,y,z) at each point when

we integrate. The value for points on the "low" end of the intervals must be subtracted,and the value for points on the "high" end of the intervals must be added. Furthermore, if

we want to evaluate the integration over the surface area of S by integrating over dS, we

need to scale the magnitude of the contribution of each point to give the appropriate

weight to each incremental region dS of the surface, because a portion of the surface thatis very oblique to the yz plane doesn't "contribute as many line segments" parallel to the x

axis as does an equal increment of the surface that is parallel to the yz plane.

Clearly the necessary scale factor at each point is the cosine of the angle between the

positive x axis and the normal to the surface at that point. Notice that this automatically

gives us the appropriate sign for each contribution as well, because (for a convex surface)the normal to the surface will have a positive x component on one end of the interval and

a negative x component on the other, which means the cosines of the respective angles

will have opposite signs. (Again, for ease of description we are assuming a convex

surface, but it's not hard to show that this restriction is unnecessary, since a given lineparallel to the x axis can have multiple segments which can be treated separately.)

Consequently, we can write our original integral as

where cos(n,i) denotes the cosine of the angle between a unit vector normal to the surface

and a basis vector in the positive x direction. Naturally, since the coordinate axes are


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symmetrical, we could arrive at analogous results with x replaced by either y or z, and ireplaced byjor krespectively.

Now suppose we are given a vector field F(x,y,z) with the components Fx(x,y,z),Fy(x,y,z), and Fz(x,y,z). Using the above results and simple additivity we have

The integrand on the left side is F, i.e. the divergence of F. Also, notice that cos(n,i),

cos(n,j), and cos(n,k) are the components of the normal unit vector n, so the integrand on

the right side is simply Fn, i.e., the dot product of F and the unit normal to thesurface. Hence we can express the Divergence Theorem in its familiar form

Several interesting facts can be deduced from this theorem. For example, if we define F

as the gradient of the scalar field (x,y,z) we can substitute for Fin the above

formula to give

The integrand of the volume integral on the left is the Laplacian of , so if is harmonic

(i.e., a solution of Laplace's equation) the left side vanishes. The integrand of the righthand integral is the normal "flux" through the surface, so we see that the integral of the

normal flux over any closed surface (in a region that everywhere satisfies Laplace's

equation) is zero. It follows that there can be no local maximum or minimum inside aregion where Laplace's equation is satisfied, because such a point would, by definition,

be completely enclosed by a surface of everywhere positive (or everywhere negative)

normal flux, making it impossible for the integral of the flux over the surface to

vanish. This, in turn, implies that if is constant over an entire closed surface (where

Laplace's equations is satisfied) then the value of is constant throughout the enclosed

volume, because otherwise the volume would have to contain a local maximum or

minimum. Combining these facts with the additivity of harmonic functions, we can

conclude that there is a unique harmonic function within an enclosed region that hasspecified values on the enclosing surface, because if 1and 2are two functions

satisfying those boundary conditions, the harmonic field 1- 2is zero over the entire

boundary, and therefore it vanishes throughout the interior as well.

If we allow non-zero charge or mass density (or sources or sinks) in the enclosed region,

such that Poisson's equation 2= -4is satisfied, then the left hand integral is the net


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charge contained within the volume, and this equals the integral of the normal flux overthe enclosing surface. This is often called Gauss' law of electrostatics, and it constitutes

one of Maxwell's equations.

Another important consequence of the Divergence Theorem can be seen by noting that

the scalar quantity nat any given point on the surface equals the partial derivative

/n where n is the displacement parameter in the direction normal to the

surface. Furthermore, if the surface in question is a sphere - which we can assume

without loss of generality is centered at the origin - then the normal displacementparameter n is equal to the radial parameter r. So for such a spherical surface we have,

Recall that, in terms of spherical coordinates r, , and , where is latitude (zero at the

North Pole) and is longitude (zero on the positive x axis) the basic line element in spaceis,

so an incremental change dat constant r and corresponds to a change ds = r d,

whereas an incremental change dat constant r and gives ds = r sin() d. Hence the

surface element in terms of and is dS = r2

sin() dd, so the preceding integral canbe written as,

Applying Leibniz'sRule for the derivative of an integral, and multiplying by 4/4, thisbecomes,

The quantity in the square brackets is just the mean value of (r,,) on the surface of thesphere, and this equation shows that the derivative of the mean value with respect to the

radius r vanishes, so the mean value of on the surface of a sphere centered at any fixed

point is independent of the radius. Considering the limit as r approaches zero, it's clear

that the mean value of a harmonic function on the surface of a sphere (of any radius) is

equal to the value of at the center of the sphere.


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SOME INTERESTING FACTS OF DIVERGENCE THEOREM:

Let's say I have a rigid container filled with some gas. If the gas starts to expand but the

container does not expand, what has to happen? Since we assume that the container does notexpand (it is rigid) but that the gas is expanding, then gas has to somehow leak out of thecontainer. (Or I suppose the container could burst, but that counts as both gas leaking out of the

container and the container expanding.)

If I go to a gas station and pump air into one of my car's tires, what has to happen to the air

inside the tire? (Assume the tire is rigid and does not expand as I put air inside it.) The air inside

of the tire compresses.

These two examples illustrate the divergence theorem (also called Gauss's theorem). Recall that

if a vector field Frepresents the flow of a fluid, then the divergence of Frepresents theexpansion or compression of the fluid.The divergence theorem says that the total expansion of

the fluid inside some three-dimensional region Wequals the total flux of the fluid out of theboundary of W. In math terms, this means the triple integral of divFover the region Wis equal

to the flux integral (or surface integral) of Fover the surface Wthat is the boundary of W(withoutward pointingnormal):

WdivFdV=WFdS.

I hope that this makes sense intuitively from the above two examples. In the first example, the

gas expanding meant divF>0everywhere in W, the inside of the container. Therefore, the net

flux out of W,WFdS, must also be greater than zero, i.e., the gas must leak out through the

container walls W. In the second example, by pumping air into the tire W, I insisted that the netflux out of the tire,WFdS, must be negative (since there was a net flux \textbf{into} the tire,and we are assuming an outwardpointing normal). By the divergence theorem, the total

expansion inside W,WdivFdV, must be negative, meaning the air was compressing.

Notice that the divergence theorem equates a surface integral with a triple integral over the

volume inside the surface. In this way, it is analogous toGreen's theorem,which equates a line

integral with a double integral over the region inside the curve. Remember that Green's theoremapplies only for closed curves. For the same reason, the divergence theorem applies to the

surface integral

SFdS

only if the surface Sis a closed surface. Just like a closed curve, a closed surface has no boundary. Aclosed surface has to enclose some region, like the surface that represents a container or a tire. In other

words, the surface has to be a boundary of some W(i.e., S=W), as described above. You cannot usethe divergence theorem to calculate a surface integral overSif Sis an open surface, like part of a cone
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or aparaboloid.If you want to use the divergence theorem to calculate the ice cream flowing out of a

cone, you have to include a top to your cone to make your surface a closed surface.

APPLICATIONS OF ELECTRODYNAMICS IN DIVERGENCE THEOREM:

The way to define a three-dimensional delta function is just to take the product of three one-dimensional functions:

The integral of this function over any volume containing the origin is again 1, and the integral ofany function of is a simple extension of the one-dimensional case:

In electrostatics, there is one situation where the delta function is needed to explain an apparent

inconsistency involving thedivergence theorem.If we have a point charge at the origin, the

electric field of that charge is

According to the divergence theorem, the surface integral of the field is equal to the volume

integral of the divergence of that field:

where the integral on the left is over some closed surface, and that on the right is over the volume

enclosed by the surface. In electrostatics, the integral on the rightevaluates to the total charge

contained in the volume divided by

Now for the catch. If we calculate (in spherical coordinates) for the point charge, we get,since only the radial component of the field is non-zero:
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At this stage, we might be tempted to say that the derivative is zero (since the derivative of any

constant is zero), but the problem is that at we also have a zero in the denominator, so wehave the indeterminate fraction of zero-over-zero. Thus although it is true that

everywhere exceptthe origin, we know from the divergence theorem that

so we must have

and

These two conditions can be satisfied if

applications of divergence (mini project)

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