volume integral

8/12/2019 Volume Integral

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Vector Calculus where S is the surface bounding a regionR in space, satisfying the condildivergence theorem. We write

Where a is a constant vector and apply d ivergence theorem to the vector functionfThus, we have

But

JJ(a x F ) . 3 =JJJdiv (a x F) dV( a x ~ ) - ; = a - F x n

and d i v ( a x F ) = V. ( a x F ) = - a . V x F (13.20)

From equations (13.18), (1 3.19) and (1 3.20), we get

Since a is an arbitrary vector, thus relation (1 3.21) yields

Th us we have proved relation (1 3.16)To prove relation (1 3.17), let us write

Where a is a constant vector and is a scalar function. Applying Gauss s theorem to f,we have

Since vecto r a is arbitrary, thus w e obtain

which is relation (13.17).

3 3 3 Integral Definitions of Gradient Divergence and Cu rlWith the h elp of the results obtained abov e, we can state the follow ing integralexpression s for grad@ div F and Curl F Let S be a closed surface bounding a volum e


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V in space, @ be a scalar function andF be a vector function defined at every point ofV. Then we have IS IS

dumc lnt grnl

div = LtS

v b o

and Curl = Lt Sv 0

These ex pressions ar e often adopted as definitions of the differential operators. Itbecomes particularly s imp le to prove the divergence and Stoke s theorem starting fromthese definitions.

13.3.4 Physical Interpretation of Divergence TheoremWe co nsider the steady flow of an incomp ressible fluid of constant densityp = 1 Sucha flow is determined by the field of its velocity vectorq. Let S be any closed surfacedrawn in the fluid, which is enclosing a volumeV. By Gauss s Divergence Theorem,

Now q . n is the component of the velocity at any point ofS in the direction of theoutward drawn normal. Thusq n denotes the amount of the fluid that flows out inunit time through the elementW Hence, the lefi hand sid e of equation 1 3.22) denotesthe amount of fluid flowing acr oss the surface S in unit time from the insid e to theoutside. This amount may be positive, negative of zero.Now the total amou nt flowing outwards must be continuously supplied so that insidethe region w e must have sou rces producing fluid. We know that divq at any pointdenotes the am ount of fluid p er unit time per unit volume that goes through ally point.Thus, div q may be through of as thesource intensity of the incon?pressible fluid atany point P Hence, the right hand side of equation (13.22) denotes the amount o ffluidper unit time supplied by the source wit in S . Thus the equality in (13.22) appearsintuitively evide nt.

We know that in body heat flows in the direction of decreasing temperature. Physicalexperiments show that the rate of flow is proportional to the gradient of thetemperature. Th is means that the velocity of the heat flow in a body is of the form

where U x,y,z) s the temperature, t is the time and k is called the thermal conductivityof the body in ordinary physical circumstancesk is a co~s tan t .Using this information

and Gau ss s divergence theorem, we now set up the mathematical model of heat flow,the so called heat-equation.

13.3.5 Modelliig of Heat Flow

Let R be a region in the body and let be its bounding surface. Th en the heat leavingRper unit time is

where, q q (; is the comp onent of in the direction of the outward unit normalvector of S From the relation (13.23) an d the divergence theorem, we obtain

On the o ther hand, the total amount of heatH in R is


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wh ere the constan t is the specific heat of the material of the body andp is the densityass per unit volum e) of the material. Hence, the time rate of d ecrease ofH is

- - auaH J J J C J z d y d z .at R

Th is rate of decrease of total heat must be equal to the abov e amoun t of heat leavin gR.Thu s from equation (13.24) we have

auJ J J o - d Y d z = - k J J J v 2 U ~ XY d z .at.

Sin ce this holds fo r any regionR in the body, thus the integrand (if continuous) mustbe zero every where , i.e.,

- c 2 v 2 u , wherec

2 kat U

Th e ab ov e partial differential equation is called the heat equation an d it isfundamental for heat conduction.

~

We next take up a basic cons eq ue ~~ cef divergence theorem, k no w ~l s Green stheorem.

13.3.6 Green s Theorem and Green s Formula

Let and g be scalar functions such that = grad g sa t is fi es the as su p t io ~~ sf thedivergence theorem in some regionR. Then

div = div (fgrad g) = v grad grad g.

Also u .n=n .Cfgradg)=f n .g radg)

Th e exp ressio qn grad g is the directiol~al erivative ofg in the direction of outwardnormal vector n of the surface S in the divergence theorem. If we denote this derivative

by * he11 the fo nnu la in the divergence theorem becom esan'

which i s cal led Green s F i r s t Formu la o r the f i r s t fo rm of Green s Th eore m.

Interchanging and g, we call obtain a similar formula. Sub stractil~ ghe two Green sfirst Formulae, w e shall obtain,

which i s cal led Green s Second Formula o r the second fo rm of Green s Theo rem.

In Green s Theorem, we have assumed that and,g are two colltinuously differentiablescalar point fu nctio ~l uch thatV fand g are aso contiiluously differentiable.

In particular, if we take

where I; is the position vector of any point relative to a fixed po int0 ithin the regionR, then g is twice continuously differentiable scalar function except at0.


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We surround y small sphere of radiusE. Let S1 be the surface of this sphere andR1 Volume Integral

be the region bounded byS and S1

We app ly G reen s Theorem to the region R1, enclosed by surface S and S1 and takingthe limit as E -- 0, we can obtain

This result is known asGreen s Formula.

Anothe r important app lication of divergen ce theorem is in terms of a basic property ofsolutions of Lap lace s E quation, which we take up next.

13.3.7 Basic Property of Solutions of Laplace s EquationConsider the formula in the divergen ce theorem, namely,

Let us assum e thatf is the gradient of a scalar function, say,f grad 7P Then

div f div (grad a) v27P

F u r t h e r , f , = f . n = n . g r a d c P.

Since the right hand side o f f , represents the directional deviative of@ in the outwarda7P

direction of S i t may be denoted as . Then formula (13.27) becomesan

Obviously this is the three-dimensional analog of the formula obtained in 12.4.3 (iii).

Taking into account the assumptions under which the divergence theorem holds, w e

imm ediately obtain, for m above, the following result:Theorem 1 :

Let @(x,y z be a solution ofLaplace s Equation

in some domain D and suppose that the secondpartial derivations of 7P arecontinuous in D Then the integral of the normal derivative of the jknction a overany piecewise smooth closed orientable surface S in D is zero.Now if we suppose that the assumptionsi n above result are satisfied by7P and is zeroeveryw here on a piec e wise smooth closed orientable surfaceS in D then puttingf = g in fonn ula (13.25) in the first form of Green s Theo rem (Section 13.3) and

denoting the interior o f S by R, we get

J J ( g r a d @ ) . ( g ~ d @ ) m = J J J i p ~ d d ~ ~ ~ ~ = ~R

Since by assumption lglad@ is continuou s inR and on S and in non-negative, it

must be ze ro everywhere inR. Hence, @ 0 and @ is constant in R an d,ax a y . az

because of continuity, equal lo its value nS. Thus,Theorem 2 :

If a function @ x, y, 2 satisfies the assumptions of Theorem and s zero at all pointsof a piecewise smooth orientable surjice S 2 D then it vanishes identically in regionR bounded by S.

Theorem 2 has an im portant conseque nce in that it gives the uniqueness theorem fo rLaplace s Equation, which can be statedas below


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Vector Calculus

Let O be a solution of Laplace s Equation which has continuous secondpartia lderivatives in a domain D, and let R be a region in wliich satisfies the assumptionof the divergence theorem. ThenO s uniquely determined tn R by its value on thebounding surface S of R.

Let us consider the following example.

Example 5 :

Show that two functions which are harmonic in a region enclosed by a surfa ce andtake on the sam e values at any point of the surface are identical.Solution :

Let O nd Y be two harmonic functions in a region bounded by a surfa ceS and letO = Y at every poi~ lt.On applying Green's divergence theorem to the function8 V 8 we obtain

Let us take 8 = O Y so thatv2 = v2O V ~ Y at everypoint of Vand8 ateverypoint ofS.

8 is constant.As 8 is zero at every point ofS thus 8 is z ro everywhere.

Let us take another example.

Example 6 :

If F = grad O nd v2O = 4 n p prove that

Whem the symbols have their usual meaning and as su ~n pti o~ lsf divergence theore111are satisfied.

Solution :

From divergence theorem, we have

= div (gradO Vv

You may now try the following exercises.

Let F = M(x,y, z ^ N(x,y, 2 3+ P(x,y, z 2 be a vector field whose c omponentsM N nd P are continuous and have continuous seco nd partial de rivatives of allkinds. Sho w that

(curl F =

for any surface to which the divergence theorem applies.


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[Hint : By direct comp utation, sho w that div CurlF) 0 and then apply divergellcetheorem to CurlF)]. -

Let S be the spherical capA +yZ + 2aZ, z r a together with its base +y2 sz = a. Find the flux of F = xz i yzj^ y2 ou tw ad through by applying thedivergence theorem.

[Hint Flux= SF ]

If q x, y , z) is the velocity vector of a differentiable fluid flow through a regionD inspace, p x,y, z, t is the density of the fluid element at each point x, y,z) at time t,then taking F - pq and u sing devergence theorem obtain the continuity equation ofhydrodynamics in the farm.

div (pq) = 0

In Unit 11 Sections 11.6 and 11.7.3, we had defined solenoidal and irrotational vectorfields as divergence free and crul free vectors respectively. We shall revisit theconcepts of so lenoidal and irrotational vector fields in the next section and express thecondition s for such vector fields in terms of integrals of vector fields and show that thetwo approaches are equivalent.

13 4 SOL ENO IDAL AND IRROTATIONAL VECTOR FIELDSREVISITED

b S ec ti o n 11.6, we had defined a solenoidal vectorF in a region if for all points in thatregion

In Section 13.3 abo ve, we have given an integral definition of divergence of a vectorfield F as

div F-

L tv -0At any point in region of volumeV bounded by surfaceS.

Volume otcllrpl


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veawcblcuh

Now if div 0, then

Also from divergence theorem, we have

v

Now if div 0 hen above result yields

SF.d s oS

across every closed surface S.

AISOJJF dS epresents flux of vector point function in a simply connected region

V bounded by S.

Thus we can define Solenoidal Vector Field asA vector point function is said to be solenoidal in a region of its flux across every

closed surfoce on that region is zero i.e. JJ S 0.S

f is a solenoidal vector then there exists a vector point finction f such that

In Section 11.7.3 we had defined a vector point function to be irrotation whencurl or there exists a scalar function O such that

By Stoke's theorem, we have

f F dr =J (curl FS

Thus, when curl is zero, we get

~ F . = o

The above relation provides another definition of an irrotational vector. We may nowdefine an irrotational vector asA contirtuous vector point function is said to be irrotational if its circulation alongevery clbsed contour C in the region i s zero.

This above definition can be useful in determining the scalar point function such that

being a given irrotational vector function. In this regard we have to remember that

A being any fixed point.Let us take some examples to illustrate the above theory.Example 7

Show that the vector point function

i w i r r n t t i n m l and find the m m s n o n d i n ~ calar function such that a


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Solution

We may verify that curl = 0.V d u m c Integral

=

To determine a e may take fixed point as 0(0,0,0) and general point P as (x, y, z).Then

Here

~ 0 . 0 ) ~ y. 0 )= xy sin z + cos x

~,o.o) Ixysinz+y2z 4 , o )Ixysinz+y2z

=cosx- +xysinz+y2z

curl =

If we omit the constant 1, then

?*1 j k;a a aax ay az

= V(cos x xy sin z y2z).

xampleIf = 0, z) (z x) + (x y) be a continuous vector point function, verify thatit is solenoidal and find the function f such that = curl

SolutionHere,

~f = cur^^letf=fl f+f2j^+h k;We suppose that f l = 0, then using expression of curl, we get

Let us take xo - 0, yo = 0Now

and\

Hence


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Vector Calculus

The general form of f; where Curlf, is

where g is any scalar point function.

You may now try the following exercises to access your knowledge.m

E 9

Show that the vector function

(sin y +z cos x)i + (x cos y sin z)3+ 0 cos z + sin x)k

is irrotational and find the corresponding scalar @ such that V a

0

Show that the following vector functions are solenoidal and find the function fsuch that curl f

i )F =xb-z)i + y(z-x)j t (x-y)ki i ) F = y z i + t x j + x y k

13 5 SUMM RY

We will now summarize the results of this unit.

Iff (x, y, z) is continuous in a domain containing R and R is bounded by

finitely many smooth curves, then, as x, y, all approach zero with

approaching infinity, then limit of the sum S - 2 (xk, yk, zk) AVk is calledk - 0

the triple integral of f (x, y, z) over the region R and is denoted by

Properties of triple integral are


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Th e volume of a regionR in 3dimensionsis given by

I nVo l u m e o f R = Lt T ] Av ~ S S S ~ VTo evaluate a triple integral, we divide the region& into ele m en klycuboidsby planes parallel to say zm or di m te plane and if z zl(x,y) and q( x, Y)are the low er and upper surfaces andA is the base of these columns and if isbounded by the curves

Y YAX), Y Y~(x), a b,

then triple integ rallJ (x, y,z)dV may be w ritten a s repeated integrals asR

Th e various physical quantities of any object in 3d ime nsio ns having variabledensity p(x, y,z which c an be evaluated with the help of triple integration are

R

First Momen ts about Coordinate Planes

Centre of Mass

Mom ents of Intertia (Second mome nts)

and 1, -JSJr2 dVR

where r x,y, z) distance of (x, y, z) to lineL.Radius of G yration about a Line

Transform ation of Volume Integrals into Surface Integrals and conversely isbased on Gauss Divergence Theorem , according to which

I ~ Re a closed bounded region inspacewhoseboundary pi wissmooth orfenfable surfaceS and if u x,y z is avector fldioq

wj jCL -d h ~ e ontinuousfirst p a ~ i a l erivativesin somedonu;al--R, then

Volume I n t e


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Where ; s the unit outward normal to the surface bounding the region R,* Under the conditions of divergence theorem,

I ~ F X -$JScurl dVS R

and JJ* ^ JJJv 9 vS R

*Integral definitions of gradient, divergellce alld curl are

JSS

grad@= Lt0 v

S S F - d sd i v F = L t

v 0

J J F X ~ S

and curl = LtS

v 0

v* We can physically interpret divergence theorem asThe amount of fluid flowing across the su rf ac es in unit time from inside tooutside equals the amount of fluid per unit time supplied by the sources withinS

*If 4 k grad U is the velocity of heat flow in a body, the11 using divergencetheorem, heat-flow equation is

u - c2v2u1at

kwhere c2 = - where rr is the specific heat of the material of the body and p is

ofthe density of the material.

* Green's Theorem statesIff and ga re two continuous functions, possessing continuous partialderivatives andS is the bounding surface of a region R satisfying conditions ofdivergence theorem, then

Sfl f v2g 6 1) v = s J ( f v RV ). dSR

* If 9 e a solution of Laplace's equation which has co~ltinuous econd partialderivatives in a domain D, and if R be a region in D satisfying the assumptionsof the divergence theorem, the11 is u~ ~ique ly etermined in R by its values onthe boundry surface S f R.

vector point function is said to solenoidal in a region if its flux across

every closed surface S in that region is zero, i.e.,$JF dS = 0

A continuous vector point f ul lc ti o~ ~F s said to irrotatio~lal f its circulatioll

along every closed center C in the region is zero, i.e., dr = 0.

SOLUTIONSIANSWERS

The gjven volulne is symnetrical about the plane XOZb 0 - IJ~erefore 7 = 0T~ find? we may consider only half the volume standing over the semicircleenclosed by

P


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p(m - mx) dy dx

n I 22(2u)'JO sin cos28 de

- on putting x = 2 a sin2 82 ( ~ ) 3 : 1 2 s i d e cos20d0

2a Y ~ a r x L

p dzdy x

V k x Z 1-p(n2x2 - m22) y x

- x o[v o - x

p(m - mx) dy dr

= n m) x expression o t ?-

2Region is the volume enclosed by the planes

~ 1 0 ,= a , y = O , y = x , z = O and z = x + y.

Here

t . J Y \ v x

olume Integral


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Vector Calculw

E .For the positive octant of the sp her eg + y 2 i a2, the limits of integration areX - O t o x = a , y = O t o y = a , z = O t o z = a .

EHere

Also volume of sphe reg +y 2 a2 is 4x2 3).

: Outward unit nonnal to S, calculated from gradient of f x ,y, z s

2 2+ , I+ xin+ y j 2~ q 2 + y 2 + Z a

and

a2= dS , s ince on the s u r f a c e o f ~ : ~ + ~ * + i ? = a ~a

From 1) and 2), divergence theorem is verified.


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, By divergence theorem,

Here F = 4 2 - 5 ' 3 + z

: div F 4x) - 2 y ) 2)ax ay

= 4 - 4 y +

Here region o f integration is + y 2= 4, z and z

ow

a p a iu a p ivand d iv cur lF) - - a x ( a y-- - y ( a =- - ~ J + Z Z - ~


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Vector alculus By divergence theorem

(Curl F I dS div Cud F v- 0 (using result(1) above)R

Hence the result.

7.Flux ofF outward through

By divergence theorem,

S S

Here

: div F - X Z - - yz) v2)ay

Hence ~ ~ u r : - d i v ~ d ~ - ~ d v =.R R

8

Com ider a volume Vcontained in a simply connected surfaceS. Let p(x, y, z) be th efluid density at any point P(x, y,z of the fluid in Vat tim et.Let e unit outward normal at element 6S of atP.

Let (a):Let q be the fluid velocity at element 6S atP.Th e equation of continuity represents the con sewa tion of mass, i.e., mass of fluidflowing out ofS per unit time equals the decrease of m ass withinS per unit time inthe ab sence of sources and sinks within V.Now mass of fluid flowing out o f S per unit time

= P q . d sS

and decrease ofmass withinS per unit time

: Continuity equation yields

* j ~ ~ d i vpq) dV - I l using divergence theorem)v V

Thi result is true for all volumes V.Thus at any point of the fluid, we have

a div (p qj 0.at


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Here

Volume ntegral

F = (siny + z c o s x ) i + x c o s y + s i n z ) j + ywsz+sinx)k

: Curl F =

Hence F s an irrotational vector point function.To find such that F = V a, we may take fixed point as 0(0,0,0) and general p i n t

t P as (x, y, z). Then

(ws)+{x,y,o) y COS z sinx) dz

x,O,O) xy,O)= xsiny+zs inx o,o,a ( X sin y y sin z

x4.0)

= xsiny +ysinz+zsinx.

Hence F = V(xsiny+ysinz+zs inx)10

(i) Here

F = - z)P+ y(z - x)I+ z x y)f;

: div = x(y - z)] -b(z - x)] z(x y)]ax ~ az-0 , -z)+(z-x)+(x-y)=O

Hence vector function F s solenoidal.

I ~ FCurlx letf=fl E+hj +fi fSuppose that l = 0, then using expression for curl, we get

Now


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(ii)

and ~ ( v , i ) = f ~ d y - ~ .0

Hence h ^+j +f,

\

The general form off, where F curlfj is

Where g s any scalar point function.

Answer

= -2y:+ y2z 2 2 ) vg2


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PPENDIX 13 1

Proof of Gauss Divergence TheoremLet S be such a surface that a line parallel to z-axis meets it in two points only. Denotethe lower and upper positions of S by S1 and S2 and let their equations be z = fi(x, y andz 2(x, y respectively.

We denote the projection of S on xy-plane by Ad -nIen

Now for the upper position S2

dxdy cosy2 S2= k-rids

and for the lower position S

dxa'y=mY1dS1 = - k . n

a negative sign being taken as the normal to dSI makes an obtuse angle x - with .

Substituting in (I), we get

=JJu3 -;

volume integral

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