PHY 301: Mathematical Methods ICurvilinear Coordinate System

(10-12 Lectures)

Dr. Alok Kumar ∗

Department of Physical SciencesIISER, Bhopal

Abstract

The Curvilinear co-ordinates are the common name of different setsof coordinates other than Cartesian coordinates. In many problems ofphysics and applied mathematics it is usually necessary to write vectorequations in terms of suitable coordinates instead of Cartesian coordi-nates. First, we develop the vector analysis in rectangular Cartesiancoordinate to see the fundamental role played by the vector-valueddifferential operator, ~∇. All objects of interests are constructed withthe del operator ~∇ - the gradient of a scalar field, the divergence ofa vector field and the curl of a vector field. Later we generalize theresults to the more general setting, orthogonal curvilinear coordinatesystem and it will be a matter of taking into account the scale factorsh1, h2 and h3. For the most general coordinate transformation wehave to consider the tensor analysis. Rectangular Cartesian coordi-nate is a special case of the orthogonal Curvilinear coordinate system,what we mean is h1 = h2 = h3 = 1. Intuitively, the scale factor is1-dimensional version of the Jacobian and we encounter the Jacobianfirstly while handling the multiple integral.

Please do not take this lecture notes at the face value, verify and check ev-erything and spot the conceptual mistakes etc. to improve it.

∗e-mail address: alok@iiserbhopal.ac.in

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The aim of vector analysis (calculus) to have the notion of the differentiationand the integration in more than one dimension. The derivative of a functionin one dimension gives the rate of the change of the function with respectto an independent variable. In more than one dimension, there are infinitelymany direction and the derivative of a scalar function φ(x, y, z) will dependon the chosen direction. In short, the question “how fast does φ(x, y, z)vary?”has an infinite number of answers, one for each direction we mightchoose to explore [2]. Geometrically, the scalar function φ(x, y, z) representsa family of surfaces and on a particular surface the value of the scalar func-tion does not change. At a point P (x, y, z) on a surface, there is a uniquenormal direction perpendicular to the surface and the directional derivativealong the normal direction is called the gradient of the scalar function,

grad φ =dφ

dnn (1)

where n is the unit vector along the normal direction. This definition is givenindependently of any coordinate system. The gradient of a scalar functionis the rate of space variation along the normal to the surface on which itremains constant [5]. With the rule of partial differentiation,

dφ =

(∂φ

∂x

)dx+

(∂φ

∂y

)dy +

(∂φ

∂z

)dz. (2)

We can rewrite the equation (2) as

dφ =

(∂φ

∂xi+

∂φ

∂yj +

∂φ

∂zk

).(dxi+ dyj + dzk) (3)

= (~∇φ).(d~l).

The vector ~∇φ = (i∂xφ + j∂yφ + k∂zφ) is the gradient of the scalar func-tion φ(x, y, z). This definition of the gradient is with respect to rectangularCartesian coordinate system. Both definitions of the gradient of a scalarfunctions equation (1) and equation (3) are equivalent. In equation (3) dφ

is a scalar and d~l is a vector and therefore ~∇φ must be a vector. Like anyvector, a gradient has magnitude and direction.

dφ = (~∇φ).(d~l) (4)

= ‖~∇φ‖‖d~l‖ cos(θ).

2

In equation (4), the change in dφ will be maximal for θ = π2

for fixed ‖d~l‖i.e. when d~l points in the direction of ~∇φ for fixed ‖d~l‖. Therefore we state,

the gradient ~∇φ points in the direction of the maximum increase of the func-tion φ. From equation (1), it is obvious that the direction of the gradient isthat of the normal to the surface given by the φ(x, y, z) = c . This can betaken as geometrical interpretation of the gradient of a scalar function. Fora point P (x, y, z) to be stationary point, ~∇φ(x, y, z) = 0.It is to be noticed that in the definition of the gradient of a scalar func-tion ~∇φ(x, y, z) = i∂xφ(x, y, z) + j∂yφ(x, y, z) + k∂zφ(x, y, z), the formal ap-

pearance of the del operator ~∇ is very crucial. Now ~∇φ is a vector quan-tity, φ(x, y, z) is a scalar quantity and therefore the differential operator~∇ ≡ ∂

∂xi+ ∂

∂yj+ ∂

∂zk must behave like a vector quantity (like Quotient rule).

It is not a vector in usual sense but it is a vector-valued differential opera-tor and hungry to differentiate [1]. Anything comes behind the ~∇, get justmultiplied ordinarily and anything comes in front of it get differentiated.So the ~∇ mimics the behaviour of an ordinary vector by acting upon anyfunction and differentiating it. This operator play a very significant role inthe whole vector calculus. ~∇ and ∇2 are invariant under rotation i.e.i∂x + j∂y + k∂z = i′∂x′ + j′∂y′ + k′∂z′ . Just imagine, with the del operator ~∇we have the power of vector algebra in one hand and the power differentialcalculus in other hand.How to prove the equivalence of two definitions of the gradient of a scalarfunction, equation (1) and equation (3)?Let us consider the scalar function in three dimensions, φ(x, y, z) = x2 +y2 +z2 = c. Obviously it represents a family of spheres centred at the (0, 0, 0)with the radius

√c and on a particular sphere the value of the scalar func-

tion φ(x, y, z) = x2 + y2 + z2 = c remains constant i.e. c will be a constanton a particular sphere. More generally, any scalar function φ(x, y, z) = crepresents a family of surfaces with different values of c. Let us consider twonearby surfaces, φ(x, y, z) = c and φ(x, y, z) = c+ ∆c with the origin chosenat the point O. P is a point on φ(x, y, z) = c and P ′ on φ(x, y, z) = c+ ∆c.~OP = ~r, ~OP ′ = ~r + d~r and ~PP ′ = d~r. Let the unit normal vector n at

the point P makes an angle θ with ~PP ′ = d~r. Hence dn = n.d~r = dr cos θ.Therefore, from equation (1),

dφ =dφ

dndn =

dφ

dnn.d~r = grad φ.d~r. (5)

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We compare equation (3) and equation (5) to get

grad φ =dφ

dnn = i

∂φ

∂x+ j

∂φ

∂y+ k

∂φ

∂z= ~∇φ (6)

and hence the both definitions are equivalent.The vector field is defined as the set all unique vectors corresponding to eachpoint (x, y, z) ∈ R3. Needless to mention the pivotal role of the del operatorin the study of vector calculus. What are the quantities of interest for thestudy the vector field? In Newtonian mechanics we study the motion of anyrigid body by breaking its motion two parts : the translational motion of thecentre of mass and the pure rotational motion in the centre of mass frame.Similarly we analyse the vector field - the linear content (the divergence) andthe rotational content (the curl) at a point.The Divergence of a Vector Field:The divergence of a vector field is closely associated with the notion of fluxof a vector field and we encounter this in the study Gauss law for the electricfield flux. The flux of any vector field across any small area around a pointis defined as

dφ ~A = ~A.d~S = ~A.n dS (7)

where n the outward normal unit vector. Physically it represents the flow ofthe vector through the elementary area. The total flux through the wholearbitrary small closed surface around the point is given by

φ ~A =∮S

~A.d~S =∮S

~A.n dS. (8)

The divergence of a vector field at a point of the vector field is the limitingvalue of the ratio of the flux of the vector field across an elementary closedsurface around the point to the volume of the enclosure when the volume ofthe enclosure contracted on to the point.

div ~A = lim∆τ→0

∮S~A.d~S

∆τ. (9)

This definition is independently of any coordinate system. Let us chooseRectangular Cartesian coordinate system and in this system the Divergenceof a Vector Field will have the following computational form with the defini-tion in equation (9)

div ~A = ~∇. ~A =∂Ax∂x

+∂Ay∂y

+∂Az∂z

. (10)

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Physically the divergence of a vector field at a point represents the ’sourcestrength’ of the vector field. If the divergence at a point is positive, the pointis behaving like the source, if it is negative the point is behaving like the sink.If the divergence of a vector field is zero at all points, we say the vector fieldis ’Solenoidal’ i.e. ~∇. ~A = 0 ∀ (x, y, z) ∈ R3. Intuitively, it is associated withthe outward normal components on an arbitrary surface around the pointand hence it is capturing the linear contents (w.r.t. normal direction) of thevector field. The definition in equation (9) is good motivation for the GaussDivergence Theorem.The surface integral of a vector field carried over the entire surface of a closedfigure is equal to the ’volume integral of the divergence of the vector field’.That is, ∮

S

~A.d~S =∫τ

~∇. ~A dτ (11)

where S is the surface area of the closed figure and τ is the volume of thespace enclosed by the same figure.The Curl of a Vector Field:The curl of a vector field around a point is defined as a vector along thenormal to an elementary area centered on the point and of the magnitudeequal to the limiting value of the ratio of the line integral to the area itself asthe area is contracted into that point.

curl ~A = lim∆S→0

∮ ~A.d~l∆S

n. (12)

where ∆S is an elementary area and n is the unit vector along the normal to∆S. This definition is also independently of any coordinate system. In theRectangular Cartesian coordinate system the above definition will be

~∇× ~A =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

Ax Ay Az

∣∣∣∣∣∣∣ . (13)

With the definition in equation (12) we can motivate to ’Stokes’s Theo-rem’,This theorem states that the line integral of a vector field along the closed pathis equal to the surface integral of the curl of the vector carried throughout thearea bounded by the path. That is,∮

S

~A.d~l =∫curl ~A.d~S. (14)

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If the curl ~A = 0 ∀ (x, y, z) ∈ R3, the vector field ~A is called irrotationalfield. Physically the curl of a vector field like vortex in the fluid motion andif we put a wheel suitably in a rotational fluid it will start rotating. Laminarflow is like irrotational velocity vector field.

The curl of a central field ~A(r) = f(r)r is zero, it has no vortex like structure,what makes it a conservative field. The gradient of a central scalar functioncan be calculated directly, ~∇f(r) = r d

drf(r).

Theorem : Curl-free (or “irrotational”) fields. The following condi-

tions are equivalent for a curl-free vector field ~A [2]:

1. ~∇× ~A = 0∀(x, y, z) ∈ R3;

2.∫ ba~A.d~l is independent of the path;

3.∮ ~A.d~l = 0 for any closed loop;

4. ~A is the gradient of some non-unique scalar field, ~A = −~∇φ.

Theorem : Divergence-less (or “solenoidal’) fields. The following

conditions are equivalent for a divergence-less vector field ~A [2]:

1. ~∇. ~A = 0∀(x, y, z) ∈ R3;

2.∫surface

~A.d~a is independent of the surface, for any given boundary line;

3.∮surface

~A.d~a = 0 for any closed surface;

4. ~A is the curl of some vector field, ~A = −~∇× ~W , where ~W is not unique.

The Laplacian operator is second order differential operator is defined as∇2 = ~∇.~∇ = ∂xx + ∂yy + ∂zz and satisfy the identity in three dimension,

∇2(

1

r

)= −~∇.

(~r

r3

)= −4πδ(r) (15)

and 1r

is interpreted as the Green’s function the Laplacian operator ∇2. Forproof of equation (15), see reference [2]. The equation is closely related tothe identity in 3-dimension

~∇.(rn~r) = (n+ 3)rn. (16)

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Left hand side of (16) is zero for n = −3, it is related to (15) in threedimension. In two dimension

~∇.(rn~r) = (n+ 2)rn. (17)

In (17) n = −2 makes zero. So in two dimension the Green’s function of the∇2 is ln(r) instead of 1

r[6].

∇2(ln(r)) = kδ(r) (18)

where k is the strength of the delta function to be determined. For m di-mension we conjecture

~∇.(rn~r) = (n+m)rn. (19)

Writing ∇2 = div grad = ~∇.~∇ and using grad f(r) = r ddrf(r)

∇2(rn) = div

(rd(rn)

dr

)= ~∇.

(n~rrn−2

)= n(n+m− 2)rn−2 (20)

for the dimension, m.There are many vector identities see [2, 3, 4] but two are of utmostimportance,

1. The curl of gradient of a scalar function is zero. ~∇× ~∇φ(x, y, z) = 0.Intuitively it state that the rotational content of a gradient (conserva-tive) vector field zero.

2. The divergence of curl of a vector field is zero. ~∇.(~∇× ~A) = 0. Intu-itively it state that the curl of a vector field has no linear content andhence its divergence is zero.

Helmholtz’s Theorem 1: A vector field is uniquely specified by giving itsdivergence and its curl within a simply connected region (without holes) andits normal component over the boundary.Helmholtz’s Theorem 2: A vector field satisfying ~∇. ~A = s and ~∇× ~A = ~cwith both the source and the circulation densities vanishing at infinity maybe written as the sum of two parts, one of which is irrotational, the other ofwhich is solenoidal.[4]We developed vector analysis in the rectangular Cartesian coordinate system.A Cartesian coordinate system offers the unique advantage that all three unit

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vectors, i, j and k are constants in magnitude as well as in direction. Unfor-tunately, not all physical problems are well adapted to a solution in Cartesiancoordinates. For instance, if we have a central force problem, ~F = rF (r),such as gravitational force, Cartesian coordinates may be unusually inappro-priate. Such a problem demands the use of a coordinate system in which theradial distance is taken to be one of the coordinates, that is, spherical polarcoordinates. The point is that the coordinate is chosen to fit the problem, toexploit any constraint or symmetry present in it. Then it is likely to be morereadily soluble than if we had forced it into a Cartesian framework [4]. Wegeneralise the results developed in the rectangular Cartesian coordinate sys-tem to orthogonal Curvilinear coordinate system and we see the translationof all results developed so far will be boiled down to taking into account thescale factors h1, h2, h3. We follow reference [3] for Curvilinear formulation.

Our physical three dimensional space is the continuum of physically points.In Cartesian it is represented by the set {(x, y, z) ∈ R3}. The same physicalpoint is denoted by (u1, u2, u3) in a general coordinate system. Both (x, y, z)and (u1, u2, u3) represent the same physical point P and therefore we demandboth coordinates are related via some mapping.

x = x(u1, u2, u3)y = y(u1, u2, u3)z = z(u1, u2, u3)

u1 = u1(x, y, z)u2 = u2(x, y, z)u3 = u3(x, y, z)

. (21)

We demand that the function between two sets (x, y, z) and (u1, u2, u3) shouldbe single valued and have continuous derivatives so that the correspondenceis unique. Therefore given a physical point P with rectangular coordinates(x, y, z) we can associate a unique set of coordinates (u1, u2, u3) called thecurvilinear coordinates. The set of the transformation (21) define a trans-formation of coordinates. Now there are three coordinate surfaces and threecoordinate curves given by

u1 = c1

u2 = c2

u3 = c3

coordinate surfaces;u1 = c1 &u2 = c2

u1 = c1 &u3 = c3

u2 = c2 &u3 = c3

coordinate curves.(22)

see the Fig. 1 in chapter 7 of reference [3]. If we keep one of the coordinatefixed and let other two vary, a surface of definite shape is traced. Suchsurfaces are called coordinate surfaces. Two surfaces cut to give a curvedline, is called coordinate curve. Let ~r = ix+ jy+ kz be the position vector of

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a physical point P and its position vector in terms of curvilinear coordinate is~r = ~r(u1, u2, u3). ∂~r(u1,u2,u3)

∂u1, ∂~r(u1,u2,u3)

∂u2and ∂~r(u1,u2,u3)

∂u3are the tangent vectors

along the u1, u2 and u3 curves respectively at the point P . The unit tangentvectors (e1, e2, e3) are

h1e1 = ∂~r(u1,u2,u3)∂u1

h2e2 = ∂~r(u1,u2,u3)∂u2

h3e3 = ∂~r(u1,u2,u3)∂u3

h1 = ‖∂~r(u1,u2,u3)

∂u1‖

h2 = ‖∂~r(u1,u2,u3)∂u2

‖h3 = ‖∂~r(u1,u2,u3)

∂u3‖

scale factors (23)

where h1, h2 and h3 are called scale factors. We know that the gradient ~∇u1

at the point P is along the normal to the surface u1 = c1 and similarly ~∇u2

and ~∇u3 are along the normal to the surfaces u2 = c2 and u3 = c3. The unitnormal vectors at the point P

E1 =~∇u1

‖~∇u1‖

E2 =~∇u2

‖~∇u2‖

E3 =~∇u3

‖~∇u3‖

. (24)

It is a simple exercise to show that two bases(∂~r(u1,u2,u3)

∂u1, ∂~r(u1,u2,u3)

∂u2, ∂~r(u1,u2,u3)

∂u3

)and (~∇u1, ~∇u2, ~∇u3) are dual (reciprocal) to each other, see the solved prob-

lem 15 of reference [3]. Any vector ~A can be written as

~A = Aiei = aiEi (25)

where dummies indices means sum over is understood. For orthogonal curvi-linear coordinate, e1.e2 = e2.e3 = e1.e3 = 0 and we are interested in thissystem. The most fundamental quantity in geometry is the distance betweentwo neighbour points

ds2 = dx2 + dy2 + dz3 = d~r.d~r. (26)

From ~r = ~r(u1, u2, u3) with the rule of partial differentiation

d~r =∂~r

∂u1

du1 +∂~r

∂u2

du2 +∂~r

∂u3

du3 = h1du1e1 + h2du2e2 + h3du3e3. (27)

The equation (26) can written as

ds2 = d~r.d~r = (h1du1e1 + h2du2e2 + h3du3e3).(h1du1e1 + h2du2e2 + h3du3e3)(28)

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ds2 =(du1 du2 du3

) h21e1.e1 h1h2e1.e2 h1h3e1.e3

h2h1e1.e2 h22e2.e2 h2h3e2.e3

h1h3e1.e3 h3h2e2.e3 h23e3.e3

du1

du2

du3

(29)

for orthonormal system the off-diagonal elements of the equation (29) willbe zero. Therefore in the orthogonal curvilinear system the distance element(the first fundamental form ) is given

ds2 = h21(du1)2 + h2

2(du2)2 + h23(du3)2. (30)

Comparing the Cartesian distance element as in equation (26) with the or-thogonal curvilinear distance element as in equation (30), we find the changealong u1 is by the scale factor h1 and similar story for other coordinate curve.The volume element would be given equation (34), scalar triple product. Thegradient, the divergence, the curl and the Laplacian in the (u1, u2, u3) is amatter taking into account the scale factor properly and the proof of allequations (36 -39) are given in the Spiegel [3].We need to study two most commonly used Orthogonal Curvilinear coordi-nate systems in full details.

1. Cylindrical Coordinates (ρ, φ, z) :The following coordinate trans-formation is called Cylindrical Coordinates.x = ρ cosφ, y = ρ sinφ, z = z;where ρ ≥ 0, 0 ≤ φ ≤ 2π,−∞ < z <∞hρ = 1, hφ = ρ, hz = 1.

2. Spherical Coordinates (r, θ, φ) :The following coordinate transfor-mation is called Spherical Coordinates.x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ;where r ≥ 0, 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ πhρ = 1, hθ = r, hφ = r sin θ.

It is to noted that for any orthogonal curvilinear coordinate systemwhat is important the scale factors as we can see in equations (36-39).

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References

1. Richard P. Feynman, Robert B. Leighton and Matthew SandsLectures on Physics, Volume 2 (Indian edition), Narosa PublishingHouse, (986).e

2. David J. Griffiths, Introduction to Electrodynamics (2nd edition),Prentice-Hall of India Private Limited, (1998).

3. Murray R. Spiegel, Theory and Problems of Vector Analysis (SI metricedition), Schaum’s Outline Series, (1974).

4. George B. Arfken and Hans J. Weber, Mathematical Methods forPhysicists (sixth edition), Academic Press, (2005).

5. N. N. Ghosh, Teach Yourself Physics Mathematical Physics, BharatiBhawan (Patna, Bihar), (1990).

6. B. S. Agarwal1 , Mechanics , Kedar Nath Ram Nath (Meerut, U.P.), (2000).

1All titles by him highly-recommended for undergraduate students .

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Quick to Curvilinear Coordinate System

Given a physical point P is represented by xi in rectangular Cartesian andby ui in any general coordinate system.

ui = ui(xj). (31)

InR3 two natural bases dual to each other are (∂~r(u1,u2,u3)∂u1

, ∂~r(u1,u2,u3)∂u2

, ∂~r(u1,u2,u3)∂u3

)

and (~∇u1, ~∇u2, ~∇u3).ds2 = d~r.d~r (32)

d~r(u1, u2, u3) =∂~r(u1, u2, u3)

∂u1

du1 +∂~r(u1, u2, u3)

∂u2

du2 +∂~r(u1, u2, u3)

∂u3

du3

(33)

dV = ‖(h1du1e1).((h2du2e2)× (h3du3e3))‖ (34)

h1 =

∥∥∥∥∥∂~r(u1, u2, u3)

∂u1

∥∥∥∥∥ ;h1e1 =∂~r(u1, u2, u3)

∂u1

h2 =

∥∥∥∥∥∂~r(u1, u2, u3)

∂u2

∥∥∥∥∥ ;h2e2 =∂~r(u1, u2, u3)

∂u2

h3 =

∥∥∥∥∥∂~r(u1, u2, u3)

∂u3

∥∥∥∥∥ ;h3e3 =∂~r(u1, u2, u3)

∂u3

(35)

~∇ui =eihi

‖~∇ui‖ =1

hi

Ei =~∇ui‖~∇ui‖

ei = Ei

for orthogonal system.

h1, h2 and h3 are scale factors owing to curvilinear nature of the coordinatesystem.

~∇φ =1

h1

∂φ

∂u1

e1 +1

h2

∂φ

∂u2

e2 +1

h3

∂φ

∂u3

e3 (36)

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Table 1: Common Scale Factors

Curvilinear Cartesian Spherical Cylindricalu1 x r ρu2 y θ φu3 z φ zh1 1 1 1h2 1 r ρh3 1 r sin θ 1

~∇. ~A =1

h1h2h3

[∂(A1h2h3)

∂u1

+∂(A2h1h3)

∂u2

+∂(A3h2h1)

∂u3

](37)

~∇× ~A =

∣∣∣∣∣∣∣e1 e2 e3

1h1

∂∂u1

1h2

∂∂u2

1h3

∂∂u3

A1 A2 A3

∣∣∣∣∣∣∣ (38)

∇2φ =1

h1h2h3

[∂

∂u1

(h2h3

h1

∂φ

∂u1

)+

∂

∂u2

(h1h3

h2

∂φ

∂u2

)+

∂

∂u3

(h1h2

h3

∂φ

∂u3

)](39)

where (e1, e2, e3) is a curvilinear orthonormal basis.

d ~A = (~∇.d~r) ~A (40)

Extensions of the above results are achieved by a more general theory ofcurvilinear systems using the methods of tensor analysis. All these resultsare smoothly derived in Spiegel [3].

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Suggested Assignments:

1. If ~A and ~B are irrotational, prove that ~A × ~B is solenoidal.

2. If f(r) is differentiable, prove that f(r)~r is irrotational.

3. If U and V are differentiable scalar fields, prove that ~5U × ~5V issolenoidal.

4. Show that ~A = (2x2 + 8xy2z)i+ (3x2y− 3xy)j− (4y2z2 + 2x3z)k is not

solenoidal but ~B = xyz2 ~A is solenoidal.

5. In what direction from the point (1,2,3) is the directional derivative ofΦ = 2xz − y2 a maximum? What is the magnitude of this maximum?

6. Find the values of the constants a, b, c so that the directional derivativeof Φ = axy2 + byz + cz2x3 at (1,2,-1) has maximum magnitude 64 in adirection parallel to the z-axis.

7. Verify Stokes’ theorem for ~A = (y− z + 2)i+ (yz + 4)j − xzk, where Sis the surface of the cube x=0, y=0,z=0,x=2,y=2,z=2 above xy plane.

8. Verify Green’s theorem in the plane for∮C(3x2−8y2)dx+(4y−6xy)dy,

where C is the boundary of the region defined by :y =√x, y = x2.

9. Prove that φ(x, y, z) = x2 + y2 + z2 is scalar invariant under rotationof axes.

10. Show that under a rotation

~∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z= i′

∂

∂x′+ j′

∂

∂y′+ k′

∂

∂z′= ~∇′ (41)

11. Show that under a rotation Laplacian operator is invariant.

12. Given the dyadic φ = ii+ jj+ kk, evaluate ~r.(~φ.~r) and (~φ.~r).~r. Is there

any ambiguity in writing ~r.~φ.~r. What is the geometrical significance of~r.~φ.~r = 1.

13. If ~A = xzi − y2j + yz2k and ~B = 2z2i − xyj + y3k, give a possiblesignificance to ( ~A× ~∇) ~B at the point (1,-1,1).

14

14. Verify the divergence theorem for ~A = 4xi− 2y2j + z2k taken over theregion bounded by x2 + y2 = 4, z = 0 and z = 3.

15. Show that1. ~5(1

r) = − ~r

r3

2.~5.(~r) = 3

3.~5× (~r) = 0

4..~5× (~a× ~r) = 2~a where ~a is a constant vector.

16. Problem numbers 37, 42, 46, 50, 53, 54, 55, 58, 59, 60, 67 of chapter 7of reference [3].It is always fine to solve many more problems as in reference [2, 3, 4].With the best of luck!!

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