UNIT 13 VOLUME INTEGRAL

Structure 13.1 Introduction 0bjectives 13.2 Triple Integral 13.2.1 Definition 13.2.2 Pmpenies of Triple Integrals 13.2.3 Volume 13.2.4 Evaluation of Triple integrals 13.2.5 Physical Applications of Triple Integrals 13.3 Transformation of Volume Integrals into Surface Integrals 13.3.1 Gauss Divergence Theorem 13.3.2 Consequen~sand Applications of Divergence Theorem 13.3.3 Integral Definitionsof Gradient, Divergence and Curl 13.3.4 Physical Interpretation of Divergence Theorem 13.3.5 Modelling of Heat Flow 13.3.6 Green's Theorem and Green's Formula 13.3.7 A Basic pmperty of Solutions of laplace's Equation 13.4 Irrotational and Solenoidal Vecto. Fields Revisited 13.5 Sununary 13.6 Solutions/Answers

13.1 INTRODUCTION

In the previous unit, you have studied about liile integrals, double ilitegrals a~idsurface integrals. In the process you have learnt to transform double integrals and surface integral into line integrals. You had learnt that line integral is the generalization of a single integral and a surface integral is a sort of generalization of a double integral. In this unit, we shall give another generalization of double integral called triple integrals or volume integrals. We shall first of all define triple integrals in Section 13.2, wherein we shall also give the properties and evaluation of such integrals. This section will be closed with some physical applications of triple integrals. We had made use of integral transform theorems - Green's Theorem and Stoke's Theorem - in the last unit. In this unit, we shall discuss another important intergral transform theorem, known as Gauss Divergence Theorem, which helps in the transformation of volume integral to surface intergal and conversely. Divergence Theorem has many important consequences and various applications, some of which have been discussed in Section 13.3. We had earlier discussed solenoidal vector fields al~dimtatiollal vector fields in Unit 11. With our knowledge of vector calculus, we have revisited these concepts in Section 13.4, where we have now given integral form conditions for vector fields to be solenoidal and irrotational. The summary of the results discussed in this unit is presented at the end of this unit. Objectives After going through this unit, you should be able to

+ define a triple integral, * state properties of triple integrals,

+ evaluate triple integrals as repeated integrals, * give some physical applications of triple integrals, * lean1 the method and conditions, under which a volume integral can be transformed into a surface integral and conversely (Gauss Divergence Ttieorem), * apply divergence theorem for the evaluation of surface integrals/volume integrals, * define gradient, divergence and curl in tenns of integrals, * understand the physical interpretation of divergence theroem, Volome Integral * use divergence theorem to set up heat equation, * derive Green's theorem in space and Green's formula, using divergence theorem, * lean1 the uniqueness property of solutions of Laplace Equation, and * express conditions for a vector field to be solenoidal and irrotation in terms of integrals. 13.2 TRIPLE INTEGRALS

In Section 12.1, the concept of a line integral was introduced. Section 12.2 was devoted to the properties and evaluation of double integrals, in which the integrand is a function of two variables. A natural generalization of double integrals was provided in Section 12.5 in terms of surface integrals, where though the integrand may be a function of three variable, but it is difined on a surface only and it is still evaluated as a double integral. However, there are many physical and geometrical situations, where the integrand may be a fuilction defined in a region in space and itegration may have to be carried out with respect to all the three variables involved. This gives rise to triple integrals. A triple integral is again a generalization of double integral (introduced in Section 12.2). 13.2.1 Definition For defining a triple integral, consider a function f (x, y, z) defined on a bounded region R in space (such as a solid cube, a ball, a truncated cone or the space between two concentric spheres). We subdivide the region R into rectangular cells by planes parallel to the three coordinate plai~es.The parallelopiped cells may have the dimensions of Ax, Ay and A z. We number the cells inside R in same order

AV,,AV, ,..., AV,

In each such parallelopiped cell; we choose an arbitrary point, say, (xbyb zk) in the kth Figu~13.1:Partitioningasolid with rectangular cells of volume AV. parallelopiped cell, and form the sum

n s~= 1 f (xb Yk, zk)AVk k- 1 (13.1) where AVk is the volume of kth cell.

We now arbitrarily do this for larger and larger positive integer 11 so that the lei~gthsof the edges of the largest parallelopiped of subdivision approaches zero as n approaches infinity. In this way, we obtain a sequence of real numbers S,, ,Snz,.... If we assume that I(x,y, z) is coi~tii~uousin a domain containing R and R is bounded by finitely many smooth curves, then as Ax, Ay, A z all approach zero with n approaching infinity, the sum S, will approach a limit [independent of choice of subdivisions and corresponding points (xb Yb zk)]

Lt s, =$$sf(4 Y, Z) dv, (13.2) n-rm R which is called the triple integral off (x, y, z) over the region R. It may be mentioned that the liinit in equation (13.2) may exist for some discoiitinuous functions also. 13.2.2 Properties of Triple Integrals Triple integrals share many algebric properties with double and single integrals. Writing F for F (x, y, z) and G for G (x, y, z), some of the properties of triple integrals (i) $$$K F N- K$$F N (any constant K) R R (ii) $$$(F * G) N =$$SF ~v*$$$G dv R R R (iii) $$$FNZO if FaOinR R (iv) $$$FN~$$$G~v if FZG~~R R R The triple integrals have the above properties because the sums that approxiinate the111 have these properties. Triple integrals also have a domain additivity property that proves useful in physics, mathematics as well as in engineering. If the domain R of a coi~tinuousfunction F be partitioned by smooth surfaces into a finite number of regions R1, R2, .. . Rn in space, then

$$$FN=$$$F ~v+$$JF dv+ ... +$$$F~v

R R 1 R2 R"

13.2.3 Valume Iff (x, y, z) = 1 is the constant function whose value is one, then the sums in Equation (1) reduce to n n s,,=c teAvk=cAvk k- 1 k- 1 As Ax, Ay and A z all approach zero, the cells A Vk become smaller and smaller and more numerous and fill up more of R. We, therefore, define the volume of R to the triple integral of constant function f (x, y, z) = 1 over R. Hence, n Volume O~R= ~t C A vx =$$$N n-w k-1 R The triple integral is seldom evaluated directly froin its defiilitioii as the liinit of a sum. Instead, one applies a three-dimensional version to evaluate the integral by repeated single integrations. This method is explained below. 13.2.4 Evaluation of Triple Integrals To express the triple integral as a repeated integral we divide the region R into eleineiitary cuboids by planes parallel to the three coordinate planes. The volume R may then be considered as the sum of vertical colun~nsextending from the lower surface of R, say z = z~ (x, y) to its upper surface z = z2 (x, y). The bases of these columns are the elementry area bA, which cover a certain region Ain the xy- plane when all the columns in R are taken.

P I

X Figure 13.2 : Evaluation of triple integral Thus, if we take the sum over the elementary cuboids in the same vertical column first, and then take the sum for all the columns in R, we can write equation (13.1) as Volume ImtqpL

t where (xbyb zk) is a point in the phcuboid above the rthelementary area Mr. Taking the limit when the dimensions of 6Ar and A z tend to zero, the above sum yields

Now integration with respect to z is performed fint, keeping x and y constant. .The remaining integration is performed as for the double integral. Therefore, if A is bounded by the curves

then triple integral in equation (13.5) may be written as

where the three integrations are performed in order from the innermost to the outermost. It should, however, be noted that the region A is the projection in the xy - plane of the bounding surface of the volume bounded by R. In case the lateral surface of the system reduces to zero, as in Figure 13.3, we may find the boundary of A by eliminating z between the two equations z = fl(x, y) and z = f2(x, y).

Figure 13.3 : Lateral surface reduces to zero

I This gives

an equation that contains no z and that defines the boundary of A in the xy-plane. To obtain the z-limits of integration in any particular instance, we may use a procedures as follows : We imagine a line L through a point (x, y) in area A and parellel to z-axis. As z increase, the line enters R at z - fi(x, y) and leaves R at z I&, y). These give the lower and upper limits of the integration with respect to z. The result of this integration is a function of x and y alone, which we can integrate supplying the limits as in double integrals. Let us take up an example to illustrate the evaluation of triple integrals. Example 1 : Find the volume enclosed between the two surface

I 7-S+7v2 and z-8-x2-v 2 Vector Calculus Solution : The given two surfaces intersect on the elliptic cylinder

2+3y2-8-3-y2

* x2 + zy2= 4

The volume projects into the area A (in the xy-plane) that is enclosed by the ellipse having tihe same equation. In the double integral with respect to y and x over- A, if we integrate with respect to y, holding x fixed, then y varies from - G2) / 2 (0 +qi7j7T. Then x varies form - 2 to 2. Thus, we have

312 4 j! 8 4-2 =fx- - 2 [248-2)c-4-3-)- ] dx = *s2 (4-~~)~'~dx=8x.fi 3 -2 You will recall that there are soinetilnes two different orders in which the single integratiotls, that evaluate a double integral, inay be worked (refer Section 12.3.4). For triple integrals there are sometimes (but not always) as inany as six workable orders of

The volume of solid shown in Figure 13.4 (a prism) is given by each of the followillg six integralls :

1-y 2

z 1 c) s1o o J~~-~dydzdr

Figure 13.4 :Volume of prism

13.2.5 Physical Applications in Three Dimensions The triple integrals find various physical applications in three dimensions. It can be used to find the mass of any object having variable density, its first moments about the coordinate planes, the centre of mass, the moments of inertia, the radius of gyration about any line, etc. Iff(x,~, 2) = &,.Y, Z) is the density of an object occupying a region R in space, then some useful farrnulae are summarized as follows : Mass m -JJJp dV , (p = density) Volume Intcgrrl R

First moments about coordinate planes :

Centre of mass :

Moments of Inertia (second moments) : 1, =$$$P e2+ 2)dv, 1, =JJJp (2+ *)d~,

1, =$$$P (1' + Y2) dV, ZL =$$$r2 P dV, where r(x, y, z) - distance from point (x, y, z) to line L. Radius of gyration about a line L :R = d((. We now take up an example to illustrate the physical applications of triple integrals. Example 2 : Find the centre of gravity of a solid of uniform density p bounded below by the disc R :2 +y2s 4 in the plane z = 0 and above by the paraboloid z = 4 - 2 - y2. Solution : The solid is shown in Figure 13.5. 4 By symmetry x = 0 = y. To find ? we calculate

2iT 2 Figure 13.5 - $ (4 - r dr d0 (in polar coordinates) 2 0-0 r-o

A similar calculation gives - - - 4 Vector Calculus Thus, the centre of gravity of the given solid is (x,y, z ) = (0,0, -)3 You may naw attempt the following exercises. E 1. Find the centre of gravity of the volume cut off from the cylinder 2 + y2 - 2an = 0 by the planes z = m and z - nu.

E 2.

~valuateff(* ex*' *' & dy cfx and state precisely what is the region of 000 integration.

E 3. Find the mass of a solid in the form of the positive octant of the sphere A? +yZ+ i? = a', if the density at any point is kxyz.

-. We shall now show that the triple integral of the divergence of a vector function u over a region R in space, under certain conditions, can be transformed into a surface integral of the normal component of u over the boundry S of R. This can be done by means of the Divergence Theonn, which is the three - dimensional analog of the Green's Theorem in the plane considered in Section 12.4. The divergence theorem is of basic importance ih various theoretical and practical considerations.

13.3 TRANSFORMATION OF VOLUME INTEGRALS INTO SURFACE INTEGRALS AND CONVERSELY

13.3.1 Gauss Divergence Theorem Statement : Let R be a closed bounded region in space whose boundary is a piece-wise smooth orientable surface S. Let u(x,y,z) be a vector finction, which is continuous and has continuous first partial derivatives in some domain containing R. Then

Where u,, = us is the component of u is the'imection of the outward normal of S with A respect to R and n is the outer unit nonnal vector of S. A If we write u and n in terms of components, say,

AAA A A A u==uli+u2j+u3kand n=cosai+cosp~+cosyk A where a, fl and y are the angles between n and the positive directions of x,y and z axes, respectively, then formula (13.6) takes the form

Since S is an orientable surface, then, by definition, this means

JJul cos a cir -JJul dy tiz

Then relation (13.7), with the help of relations (13.8 a,b,c), may be written as

It is clear that relation (13.9) is true, if the following three relations hold simultaneously :

For Divergence Theorem, either the region R in space is some convex region with no holes (such as the interior of a sphere, or a cube, or an ellipsoid) or R is bounded by a piecewise smooth orientable surface S. In the later case, region R has the property that any straigbt line parallel to any one of the coordinate axes and intersecting R has at most one segment (or a single point) in common with R. This implies that R can be represented in the form

where (x,y) varies in the orthogolial projection of R in the xy-plane. It may be added that the divergence theorem is also true for region R which can be subdivided into finitely many regions of the type discussed above by means of auxiliary surfaces. It will be seen that the surface integral over the auxiliary surfaces cancel in pairs, and the sum of the remaining surface integrals is the surface integral over the whole boulidary S of R. The proof of Divergence Theorem is given in Appendix 13.1 and tho~interested in proof may look up the Appendix 13.1. We shall now take an example to evaluate a surface integral by the divergence theorem. Example 3 : Evaluate

~=~~(x~d~&+~~dzdr+x~zdrdy) S Where S is the closed surface consistintof the cylinder 2 +y2= d (0 s z s b) and the circular disks z = 0 and z = (x2 + y 2-s a 2). 43olution : If we compare the given surface integral with the right-hand side of relation (13.9) above for divergence theorem, Vector Calculus Hence, if we use the symmetry of the region R bounded by S, the volume integral in relation (13.9) assumes the form -

Thus

Let y=acost = 201) 4 2-0 y-o 3 dy=-asii~tdt where y=O,t=,JC where y=a, t=O a2-y2=a2 sin 2 t

We take up another example wherein we shall verify the divergence theorem. Example 4 : Verify divergence theorem for the region

Solution : Here

and

The given region can be expressed as 1 5 p s 2, (aimular region between two spheres). Thus, throughout the region 1 s p s 2, all functions considered are continuous. Also

Thus

$$$div F dV = 0 R

On the outer sphere (p = 2), the unit normal is

Hence On the inner sphere (p l), the positive unit nonnai points towards the origin and it Volume lnttgral is given by

Hence

Thus, 1 J$F-;~s=JJ-ds=4np2(-2 P- 1 -4x. (13.1

The sum of equations (13.14) and (13.15) is the surface integral over the complete boundary of the given region, which is

-4n+4n=O,

which agrees with equation (13.13) and this verifies the divergence theorem. You may now attempt the following exercises. E 4. Verify divergence theorem for the sphere

Use divergence theorem to evaluate the surface integral J$F dS, where S F = 41 f- 2y2j^+2 and S is the surface bounding of the region x2 + y2 = 4, z = 0 and 2-3.

There are many important consequences and various applications of divergence theorem, some of which we shall take up next. 13.3.2 Consequences and Applications of Divergence Theorem The Gauss's Divergence Theorem enables us to express a normal surface integral as a volume integral. We shall now see that with the help of this theorem, we are able to express the other two types of surface integral as volume integral, i.e, we shall show that

and Vector Calculus where S is the surface bounding a region R in space, satisfying the condil divergence theorem. We write

Where a is a constant vector and apply divergence theorem to the vector function f. Thus, we have

JJ(a x F) . ;d!3 =JJJdiv (a x F)dV S R

A But (ax ~)-;=a-Fxn

and div(axF)=V.(ax F)=-a.Vx F (13.20)

From equations (13.18), (13.19) and (1 3.20), we get

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

Thus 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 tof, we have

1 R 1 Since vector a is arbitrary, thus we obtain

which is relation (13.17). 1333 Integral Definitions of Gradient, Divergence and Curl With the help of the results obtained above, we can state the following integral expressions for grad @, div F and Curl F. Let S be a closed surface bounding a volume I I V in space, @ be a scalar function and F be a vector function defined at every point of Vdumc lntcgrnl I I V. Then we have IS+ 'IS

div F = Lt S v-bo v '

S and Curl F = Lt v-0 v These expressions are often adopted as definitions of the differential operators. It becomes particularly simple to prove the divergence and Stoke's theorem starting from these definitions. 13.3.4 Physical Interpretation of Divergence Theorem I We consider the steady flow of an incompressible fluid of constant density p = 1. Such a flow is determined by the field of its velocity vector q. Let S be any closed surface drawn in the fluid, which is enclosing a volume V. By Gauss's Divergence Theorem,

h Now q . n is the component of the velocity at any point of S in the direction of the A outward drawn normal. Thus q - n denotes the amount of the fluid that flows out in unit time through the element W. Hence, the lefi hand side of equation (13.22) denotes the amount of fluid flowing across the surface S in unit time from the inside to the outside. This amount may be positive, negative of zero. Now the total amount flowing outwards must be continuously supplied so that inside the region we must have sources producing fluid. We know that div q at any point denotes the amount of fluid per unit time per unit volume that goes through ally point. Thus, div q may be through of as the source-intensity of the incon?pressible fluid at any point P. Hence, the right-hand side of equation (13.22) denotes the amount offluid per unit time supplied by the source within S. Thus the equality in (13.22) appears intuitively evident. We know that in body heat flows in the direction of decreasing temperature. Physical experiments show that the rate of flow is proportional to the gradient of the temperature. This means that the velocity q of the heat flow in a body is of the form

where U(x,y,z)is the temperature, t is the time and k is called the thermal conductivity of the body - in ordinary physical circumstances k is a co~stant.Using this information and Gauss'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 S be its bounding surface. Then the heat leaving R per unit time is

S where, q,, - q . (; is the component of q in the direction of the outward unit normal vector of S. From the relation (13.23) and the divergence theorem, we obtain

On the other hand, the total amount of heat H in R is where the constant 0 is the specific heat of the material of the body and p is the density ( ass per unit volume) of the material. Hence, the time rate of decrease of H is au - --aH - -JJJCJp z&dy dz. at R This rate of decrease of total heat must be equal to the above amount of heat leaving R. Thus from equation (13.24) we have au -JJJo -&dYdz=- kJJJv2 U~XdYdz. R at. R

Since this holds for any region R in the body, thus the integrand (if continuous) must be zero every where, i.e.,

j -c2v2u, where c 2k=- at UP The above partial differential equation is called the heat - equation and it is fundamental for heat conduction. - -~ We next take up a basic conseque~~ceof divergence theorem, know~las Green's theorem. 13.3.6 Green's Theorem and Green's Formula Let f and g be scalar functions such that u = f grad g satisfies the assuptio~~sof the divergence theorem in some region R. Then

div u = div (fgrad g) = f v2g + grad f - grad g. A A A Also u.n=n.Cfgradg)=f(n.gradg) A The expressioqn . grad g is the directiol~alderivative of g in the direction of outward normal vector n of the surface S in the divergence theorem. If we denote this derivative by the11 the fonnula in the divergence theorem becomes *an'

which is called Green's First Formula or the first form of Green's Theorem. Interchanging f and g, we call obtain a similar formula. Substractil~gthe two Green's first Formulae, we shall obtain,

which is called Green's Second Formula or the second form of Green's Theorem. In Green's Theorem, we have assumed that f and ,g are two colltinuously differentiable scalar point functio~lsuch that Vfand Vg are aso contiiluously differentiable. In particular, if we take

where I; is the position vector of any point relative to a fixed point 0,within the region R, then g is twice continuously differentiable scalar function except at 0. We surround 0 by small sphere of radius E. Let S1 be the surface of this sphere and R1 Volume Integral be the region bounded by S and S1. We apply Green's Theorem to the region R1, enclosed by surface S and S1 and taking the limit as E --, 0, we can obtain

This result is known as Green's Formula. Another important application of divergence theorem is in terms of a basic property of solutions of Laplace's Equation, which we take up next. 13.3.7 A Basic Property of Solutions of Laplace's Equation Consider the formula in the divergence theorem, namely,

Let us assume that f is the gradient of a scalar function, say, f = grad 7P. Then

div f = div (grad a)= v27P

A A Further, f,=f.n=n.gradcP. Since the right hand side off, represents the directional deviative of @ in the outward a7P direction of S, it may be denoted as -. Then formula (13.27) becomes an

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, we immediately obtain, form 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 are continuous in D. Then the integral of the normal derivative of the jknction a over any piecewise smooth closed orientable surface S in D is zero. Now if we suppose that the assumptions in above result are satisfied by 7P and is zero everywhere on a piece wise smooth closed orientable surface S in D, then putting f = g = in fonnula (13.25) in the first form of Green's Theorem (Section 13.3) and denoting the interior ofS by R, we get

JJ$(grad@).(g~d@)m=JJJip~dd~~~~=~ R R Since by assumption lglad @( is continuous in R and on S and in non-negative, it must be zero everywhere in R. Hence, -a@ = -= = 0 and @ is constant inR and, ax ay . az because of continuity, equal lo its value 0 on S. Thus, Theorem 2 : If a function @(x, y, 2) satisfies the assumptions of Theorem 1 and is zero at all points of a piecewise smooth orientable surjice S 112 D, then it vanishes identically in region R bounded by S. Theorem 2 has an important consequence in that it gives the uniqueness theorem for Laplace's Equation, which can be stated as below : Vector Calculus Let O be a solution of Laplace's Equation which has continuous secondpartial derivatives in a domain D, and let R be a region in D wliich satisfies the assumption of the divergence theorem. Then O is uniquely determined tn R by its value on the bounding 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 surface and take on the same values at any point of the surface are identical. Solution : Let O and Y be two harmonic functions in a region bounded by a surface S and let O = Y at every poi~ltS. On applying Green's divergence theorem to the function 8 V 8, we obtain

Let us take 8 = O - Y so that v20 = v2O - V~Y= 0 at everypoint of Vand 8 - 0 at everypoint of S.

* 8 is constant. As 8 is zero at every point of S, thus 8 is zero everywhere. Let us take another example. Example 6 : If F = grad O and v2O = - 4 n p, prove that

Whem the symbols have their usual meaning and assu~nptio~lsof divergence theore111 are satisfied. Solution : From divergence theorem, we have

=$$$div (grad O)dV v

You may now try the following exercises. E 6. Let F = M(x, y, z) i^+ N(x, y, 2)3+ P(x, y, z) 2 be a vector field whose components M, N and P are continuous and have continuous second partial derivatives of all kinds. Show that

$$(curl F) ;B = o S for any surface to which the divergence theorem applies. Volume Iotcllrpl [Hint : By direct computation, show that div (Curl F) = 0 and then apply divergellce theorem to (Curl F)].--

Let S be the spherical cap A? + yZ + 2 = 2aZ, z r a together with its base 3 + y2 s a', z = a. Find the flux of F = xz i^- yzj^+ y2 $ outwad through S by applying the divergence theorem. [Hint : Flux =$SF- ;&] S

If q(x, y, z) is the velocity vector of a differentiable fluid flow through a region D in space, 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 using devergence theorem obtain the continuity equation of hydrodynamics in the farm.

div (pq) + 2 = 0,

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

13.4 SOLENOIDAL AND IRROTATIONAL VECTOR FIELDS REVISITED bSection 11.6, we had defined a solenoidal vector F in a region if for all points in that region

In Section 13.3 above, we have given an integral definition of divergence of a vector field F as

- - div F - Lt S v-0 v At any point in region of volume V bounded by surface S. veawcblcuh Now if div F - 0, then

Also from divergence theorem, we have

S v Now if div F = 0, then above result yields

$SF. ds- o S across every closed surface S. AISOJJF dS represents flux of vector point function F in a simply connected region S V bounded by S. Thus we can define Solenoidal Vector Field as A vector point function F is said to be solenoidal in a region of its flux across every closed surfoce S on that region is zero, i.e., JJF. dS = 0. S If 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 F to be irrotation when curl F = O or there exists a scalar function O such that

By Stoke's theorem, we have

f F . dr =J$(curl F) . l C S Thus, when curl F is zero, we get ~F.&=o C The above relation provides another definition of an irrotational vector. We may now define an irrotational vector as A contirtuous vector point function F is said to be irrotational if its circulation along every clbsed contour C in the region is zero. This above definition can be useful in determining the scalar point function @ such that

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

D

A being any fixed point. Let us take some examples to illustrate the above theory. Example 7 : Show that the vector point function

iw irrntatinml and find the mmsnondin~scalar function such that F - V a. Solution : Vdumc Integral We may verify that curl F = 0. Here ?* 1 j^ k; a a a curl F = - - - ax ay az

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

(~0.0) (~y.0) = xy sin z + cos x + Ixysinz+y2z I 1 (~,o.o) + Ixysinz+y2z 1 (4,o) =cosx- 1 +xysinz+y2z If we omit the constant - 1, then

F = V(cos x + xy sin z + y2z).

Example 8 : A A If F = 0, - z) i + (z - x) j + (x - y) $ be a continuous vector point function, verify that it is solenoidal and find the functionf such that F = curl$ Solution : Here,

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

Let us take xo - 0, yo = 0 Now

and \ Hence Vector Calculus

The general form off; 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

A A F = (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 F = V a.

E 10. Show that the following vector functions F are solenoidal and find the functionf such that F = curl f : A A A i)F'=xb-z)i + y(z-x)j + t(x-y)k A A A ii)F=yzi+txj+xyk

13.5 SUMMARY

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 A x, A y, & all approach zero with n n approaching infinity, then limit of the sum S, - 2 f (xk, yk, zk) AVk is called k-0 the triple integral off (x, y, z) over the region R and is denoted by

* Properties of triple integral are Volume Inte <'

The volume of a region R in 3dimensions is given by n I VolumeofR= Lt T]Av~=SSS~V

To evaluate a triple integral, we divide the region& into elemenkly cuboids by planes parallel to say zmordimte plane and if z = zl(x, y) and z - q(x, Y) are the lower and upper surfaces and A is the base of these columns and if A is bounded by the curves

Y = Y AX), Y = Y~(x),x = a, x = b,

6 then triple integrallJ$ f (x, y, z)dV may be written as repeated integrals as R

n The various physical quantities of any object in 3dimensions having variable density p(x, y, z) which can be evaluated with the help of triple integration are

R First Moments about Coordinate Planes :

I Centre of Mass :

Moments of Intertia (Second moments) :

and 1, -JSJr2 P dV. R where r(x,y, z) = distance of (x, y, z) to line L. Radius of Gyration about a Line L :

Transformation of Volume Integrals into Surface Integrals and conversely is based on Gauss Divergence Theorem, according to which "I~Rbe a closed bounded region in space whose boundary pieeewise smooth orfenfable surface S and if u(x,y, z) is a vector bfldioq nd h~e !- 1 al-- -ontinuous first pa~ialderivatives in some donu;,wj,jCL

R, then ~ - Where ;is the unit outward normal to the surface S bounding the region R," * Under the conditions of divergence theorem,

I~FXa = -$JScurl F dV ' S R and JJ* n^ a =JJJv 9 dv S R * Integral definitions of gradient, divergellce alld curl are JS* grad@= Lt -S v-0 v SSF-ds divF=Lt v-0 v JJFX~S and curl F = Lt S v-0 v * We can physically interpret divergence theorem as The amount of fluid flowing across the surfaces in unit time from inside to outside equals the amount of fluid per unit time supplied by the sources within S. * If 4- - k grad U is the velocity of heat flow in a body, the11 using divergence theorem, heat-flow equation is

--au - c2v2u1 at k where c2 = -, where rr is the specific heat of the material of the body and p is of' the density of the material. * Green's Theorem states Iff and gare two continuous functions, possessing continuous partial derivatives andS is the bounding surface of a region R satisfying conditions of divergence theorem, then

Sfl f v2g - 6'1)dv=sJ( f vg - RV!). dS. R * If 9be a solution of Laplace's equation which has co~ltinuoussecond partial derivatives in a domain D, and if R be a region in D satisfying the assumptions of the divergence theorem, the11 is u~~iquelydetermined in R by its values on the boundry surface S of R. * A vector point function F 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. S * A continuous vector point fullctio~~Fis said to irrotatio~lalif its circulatioll along every closed center C in the region is zero, i.e., $ F - dr = 0. C

SOLUTIONSIANSWERS

The gjven volulne is symnetrical about the plane XOZb 0)-'IJ~erefore 7 = 0- T~ find?, we may consider only half the volume standing over the semicircle enclosed by

I 3 P. Volume Integral

p(m - mx) dy dx

nI2 2(2u)'JO sin% cos28 de - , on putting x = 2 a sin2 8 2(~)3$:12 side cos20 d0

2a Y~ar-xL p dzdy dx

L Vk-xZ 1 -p(n2x2 - m22) dy dx - Jx.o[v.o - x22 p(m - mx) dy dr

1 = $n- + m) x expression o t ?

E 2. Region is the volume enclosed by the planes

~10,x=a, y=O, y=x, z=O and z=x+y.

Here

t! ., , J+Y- pO\ dv dx Vector Calculw

E 3. For the positive octant of the sphereg +y2 + i? = a2, the limits of integration are X-Otox=a,y=Otoy=a,z=Otoz=a.

E 4. Here

Also volume of sphereg +y2 + 2 a2 is (4x2i 3).

:. Outward unit nonnal to S, calculated from gradient off (x,y, z) is

A 2(2+ A) xin+ yj+ 2 11 = ,!I+ ~q2+y2+Z= a and

a2 =-dS ,since on the surfaceof~:~+~*+i?=a~ a

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

Here F = 42- 5'3+ z2 a a a :. div F- -(4x) + -(- 2y)+ $2) ax ay

=4-4y+& Here region of integration is 2 +y2= 4, z - 0 and z - 3

Now

a ap a a aiu ap a aiv a and div(curlF)--ax(ay --- 4+- ay(a= --- ~J+Z(Z-~ Vector Calculus By divergence theorem $$(Curl F) - I; dS -$$$div (Cud F)dv- 0 (using result (1) above) S R Hence the result. E 7. Flux of F outward through

By divergence theorem,

S S Here

a a :. div F ~XZ)+ -(- yz) + $v2) - ay

Hence ~~ur:-$$$div~d~-$$$~dv=0. R R E 8. Comider a volume Vcontained in a simply connected surface S. Let p(x, y, z) be the fluid density at any point P(x, y, z) of the fluid in Vat time t. Let ;; be unit outward normal at element 6S of S at P. Let ss = (a): Let q be the fluid velocity at element 6S at P. The equation of continuity represents the consewation of mass, i.e., mass of fluid flowing out of S per unit time equals the decrease of mass within S per unit time in the absence of sources and sinks within V. Now mass of fluid flowing out ofS per unit time

=$$P q .,ds S and decrease of mass within S per unit time

:. Continuity equation yields

*j~~div(pq) dV = -$$I$dl (using divergence theorem) v V

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

a + div (pqj 0. at - E 9. Volume Integral Here

A A A F= (siny +zcosx)i+(xcosy+sinz)j+(ywsz+sinx)k

a :. Curl F=

* Hence Fis an irrotational vector point function. To find such that F= V a, we may take fixed point as 0(0,0,0) and general pint t P as (x, y, z). Then

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

(x,O,O) (xy,O) = xsiny+zsinx I(o,o,a + (X sin y + y sin z I 1 (x4.0)

= xsiny +ysinz+zsinx. Hence F = V(xsiny+ysinz+zsinx) E 10. (i) Here

F= - z)P+ y(z - x)I+ z(x - y)f; a a a :. div F = - [x(y - z)] + -b(z - x)] + - [z(x - y)] ax a~ az -0,-z)+(z-x)+(x-y)=O

Hence vector function Fis solenoidal. I~F= Curlx letf=fl E+hj'+fi f; Suppose that fl = 0, then using expression for curl, we get

Now and ~(v,i)=f~dy-~. 0 Hence f =hi^+j$+f, $ I

I \ I The general form off,'where F = curlfj is

Where g is any scalar point function. (ii) Answer : 1 1 f = -2y:+ 5(y2z - 22) + vg. 2 i APPENDIX 13.1

Proof of Gauss Divergence Theorem Let S be such a surface that a line parallel to z-axis meets it in two points only. Denote the lower and upper positions of S by S1 and S2 and let their equations be z = fi(x, y) and z - f2(x, y) respectively.

We denote the projection of S on xy-plane by A. d -nIen

A

Now for the upper position S2 A A dxdy = cosy2&S2= k-rids, and for the lower position S1 A A dxa'y=mY1dS1 = -k.n& a negative sign being taken as the normal to dSI makes an obtuse angle x - yl with &.

Substituting in (I), we get

I =JJu31 -;& I S This proves relation (13.12) of Section 13.3. Similarly we can prove relations (13.10) and (13.1 1) of Section 13.3 I This completes the pmf of relation (13.9) and Gauss Divergence Theorem.