CSK HP Agriculture University-Palampur College of Basic ...

QMS IS FORM No. QSP/7.1/01.F02

CSK HP Agriculture University-Palampur College of Basic Sciences,

Department of Physical Sciences and Languages

Course Contents & Course Outline 1.Academic Year: 2019-20 2. Semester 6

th

1. 3. Course No. Math 323 2. 4. Credit Hours 3+0

3. 5. Course Title: Vector Calculus

4. 6. Name of Instructor (s): Dr Shweta Pathania

7. Lecture/Practical Schedule:

Name of Topic/Practical No. of tentative Lectures /Tutorial

No. of Delivered Lectures /Tutorial *

Remarks*

UNIT I Scalar and vector product of three vectors 4

Product of four vectors, Reciprocal Vectors. 6 Differentiation and partial differentiation of a vector function. 7

Gradient of a scalar point function, divergence and curl of vector point function.

8

UNIT II

Vector integration: line integral, 5

surface integral, 4

Volume integral 4

Theorems of Gauss, Green and Stokes (without proof) and the

problems based on these theorems.

7

*Information to be supplied to the HOD after the completion of Course 8. Main References (Theory/Tutorial):

1 G.B. Thomas and R.L. Finney, Calculus, 9th Ed., Pearson Education, Delhi, 2005. 2. H. Anton, 1. Bivens and S. Davis, Calculus, John Wiley and Sons (Asia) P. Ltd. 2002. 3. P.C. Matthew's, Vector Calculus, Springer Verlag London Limited, 1998. 9. Examination schedule and distribution of marks Mid-Term –25%;Assignment/Tutorial-15%; End of Term – 60% .

Approved By

HOD Course Instructor(s)

Copy to: 1. The Head, Department of Physical Sciences and Languages, COBS, CSKHPKV-Palampur 2. The Dean, College of Basic Sciences, CSKHPKV-Palampur Copy to:

Dean, College of Basic Sciences,CSKHPKV,Palampur through Head, Dept. Physical Sciences and Languages,

COBS, CSKHPKV,Palampur

SCALAR TRIPLE PRODUCT

PROBLEMS:- 1) Find the volume of parallelepiped whose edges are represented by sol: Volume of a parallelepiped = = 2(4-1)+3(2+3)+4(-1-6)

= 7 cubic unit.

kjiC

kjiB

kjiA

ˆ2ˆˆ3

ˆˆ2ˆ

ˆ4ˆ3ˆ2

213

121

432

2) Show that points (4,5,1) , (0,-1,1) , (3,9,4) , (-4,4,4) are coplanar.

Sol:

kji

kjikji

AODODA

kji

kjikji

AOCOCA

ji

kjikji

ABBA

kji

let

ˆ3ˆˆ8

ˆˆ5ˆ4ˆ4ˆ4ˆ4

ˆ3ˆ9ˆ

ˆˆ5ˆ4ˆ4ˆ9ˆ3

ˆ6ˆ4

ˆˆ5ˆ4ˆˆˆ0

00

ˆ0ˆ0ˆ00

This can be represented as = = -4 (-27+3) + 6 (-3+24) + 0 = -4 (-24) + 6 (21) = 96 + 126 = 222 ≠ 0 as the result is not equal to zero so these given points are not coplanar.

318

391

064

3)Find ‘P’ such that are coplanar.

kjiC

kjpiB

kjiA

ˆ5ˆ4ˆ3

ˆ3ˆˆ

ˆˆˆ2

0

543

31

112

p

= 2 (5p+12) + 1 (5+9) + (4-3p) = 0 = 10p + 24 + 14 + 4 – 3p = 0 = 7p = -42 = p = -6

Product of four vectors

.

RECIPROCAL SYSTEM OF VECTORS

QUESTIONS

DERIVATIVE OF A VECTOR-VALUED FUNCTION

•

•

•

•

•

•

•

•

•

•

•

Partial differentiation of a vector function

Firstly , A partial derivative is the rate of change of a multi-variable function when we allow only one of the variables to change. Specifically, we differentiate with respect to only one variable, regarding all others as constants (now we see the relation to partial functions!). Which essentially means if you know how to take a derivative, you know how to take a partial derivative.

A partial derivative of a function f with respect to a variable x, say z=f(x,y1,y2,...yn ) (where the yi' s are other independent variables) is commonly denoted in the following ways:

Divergence of a Vector Point Function

Examples

Surface Integral

Any integral which is to be evaluated over a surface is called a

surface integral. Let be a single valued function

defined over a surface S of finite area. Subdivide the area S into

n elements of areas In each part , we

choose an arbitrary point . Form the sum

. Take the limit of this sum as in such a way that the

largest of the areas approaches zero. This limit if it exists,

is called the surface integral of over S and is denoted by

zyxf ,,

.,......, 2`1 nSSS S kkkk zyxP ,,

n

K

kk SPf1

)(

n

kS zyxf ,,

or

S

dSzyxf ),,( S

fdS .

kS

kP

Flux across a surface

Let s be a piecewise smooth surface and be a

vector function of position defined and continuous over S.

Let P be any point on the surface S and be the unit vector at

P in the direction of outward drawn normal to the surface S at P.

Then is the normal component of at P. The integral of

over S i.e., is called the flux of over S.

Let be a vector of magnitude dS and direction that of

. Then

zyxF ,,

n

nF .

F

nF . dSnFS

.

F

dS

n

dSndS

SS

dSFdSnF ..

Let be the angles which makes with co-ordinate

axes. If l,m,n are the direction- cosine of this outward normal,

then

Let

,,

n

cos,cos,cos nml

knjmilkjin coscoscos

kFjFiFF 321

coscoscos. 321 FFFnF

nFmFlF 321

SS

dSnF. dSFFF )coscoscos( 321

)( 321 dxdyFdzdxFdydzF S

n

F

Question 1: Evaluate where

and S is the surface of the plane in the

First octant cut off by the plane

Sol: Here

A vector normal to the surface is given by

dSnFS

.

kjxiyF 2

62 yx

.4z

kjxiyF 2

jiyx

zk

yj

xiyx 262)62(

)2(5

1

14

2

S of),,(point any at normalunit a

jiji

zyxn

RS

S

dxdzxdxdyx

dSn

jjijn

yxxxyji

kjxiyn

jn

dxdzndSn

)6(2

5

15

)6(2.F

5

1)2(

5

1.

S.on 62 as5

)6(2

5

22

5

2).2(.F

plane xzon the S of Projection theis R Where

.

.F.F that know we

S

S

1082

2780

2

9188

268

)6(42)6(2)6(2

3

0

2

3

0

4

0

3

0

3

0

4

0

xx

dxxdxxzdxxdzxx z

Question 2:

Sol:

A vector normal to the surface S is given by

5.z and 0zbetween octant first in the included 16cylinder x theof surface

theis S and3F where,. Evaluate

22

2

y

kzyjxizdSnFS

kzyjxiz 23F Here

444

22

S of z)y,(x,point at normalunit

221616

2222

2222

jyix

yx

jyix

yx

jyix

n

jyixyxz

ky

jx

iyx

We know that

dxdzxy

zx

dxdzy

xyzxdSnF

yj

jyixjn

xyzxjyix

kzyjxizn

jn

dxdznFdSnF

z x

z xS

RS

5

0

4

0

5

0

4

0

2

4.

4

)(.

4.

4.

)(4

1

4)3(.F Now

plane. xzon the S of projection theis R Where

.

..

=

.900504028

482

16

16

5

0

2

5

0

4

0

5

0

22

5

0

4

02

zz

zx

xz

dxdzxx

zx

zxz

z x

VOLUME INTEGRAL : -

Let V be volume bounded by surface S . Let

( x , y , z ) be single valued function of

position defined on V and is denoted by

∫∫∫vf·dv OR ∫vf·dv ·

If volume V divided into small cuboid by

drawing small lines similarly let to three

coordinates

dv = dxdydz ·

∴ Volume integral = ∫∫∫vfdxdydz ·

If F vector is a vector point function , then

volume integral = ∫∫∫vFdV ·

et = F1 +F2 +F3 then

Volume integral = ∫∫∫v dxdydz + ∫∫∫vFdxdydz

+ ∫∫∫vFdxdydz ·

Question 1·

Evaluate ∫∫∫E12y - 8xdV where E is the

region behind y = 10 - 2z and in front of

region in the xz-plane bounded by z = 2x , z

= 5 and x = 0 .

SOL. The region in xy-plane bounded by z =

2x , z = 5 and x = 0 .

Therefore , we have

∫∫∫E12y - 8x dV = ∫05 ∫0

½z ∫010-2z 12y -

8xdydxdz ·

∫∫∫E12y - 8xdV = ∫05 ∫0

½z ( 6y2 - 8xy )│010-2z

dxdz ·

= ∫05 ∫0

½z 6(10 - 2z)2 - 8x (10 -

2z)dxdz ·

= ∫05 [6(10 - 2z)2x - 4x2(10 -

2z) ]│0½zdz ·

= ∫05 3z(10 - 2z)2 - z2(10 -

2z)dz ·

= ∫05 14z3 - 130z2 - 300zdz ·

= (7/2z4 - 130/3z3 + 150z2)│05 ·

= 3125 / 6 ·

QUESTION 2 .

Evaluate ∫∫∫v ( 2x + y )dV , where V is the

closed region bounded by the cylinder z = 4

- x2 and the plane x = 0 , y = 0 , x = 2 , y = 2

and z = 0 .

SOL . ∫∫∫v(2x + y )dV = ∫y=02 ∫x=0

2 ∫z=04-x2 (2x +

y)dzdxdy ·

= ∫y=02 ∫x=0

2 (2x + y)[z]04-

x2 dxdy ·

= ∫y=02 ∫x=0

2(2x + y)(4 -

x2)dxdy ·

= ∫y=02 ∫x=0

2[(2x (4 - x2) +

y(4 - x2)]dxdy ·

= ∫y=02 ∫x=0

2(8x - 2x3 + 4y

- x2y)dxdy ·

= ∫y=02 [ 4x2 - x4/2 +4xy -

x3y/3 ]02dy ·

= ∫y=02 [ 16 - 8 + 8y -

8/3y ]dy ·

= ∫y=02 [ 8 + 16/3y ]dy

= [ 8y + 8/3y2 ]02

= ( 16 + 32/3 ) ·

= 82/3 .

QUESTION 3

Determine the volume of the region that lies

behind the plane x + y + z = 8 and in front of

the region in the yz-plane that is bounded by z

= 3/2√y and z= 3/4y .

SOL .

V = ∫∫∫E dV = ∫∫D [ ∫08-y-zdx ]dA ·

= ∫04 ∫3y/4

3√y/28-y-zdzdy ·

= ∫04 ( 8z - yz - ½z2 )│3y/4

3√y/2dy ·

= ∫04 ( 12y½ - 57/8y - 3/2y3/2 +

33/32y2dy ·

= ∫04 12y½ - 57/8y - 3/2y3/2 +

33/32y2dy ·

= ( 8y3/2 - 57/16y2 - 3/5y5/2 +

11/32y3 ) ·

GAUSS DIVERGENCE THEOREM

vector.normaldrawn outwardunit theis n wheredvFdiv dSn.FThen

V.region a enclosing surface closed

a is S andfunction point vector abledifferentiy continuesl a is F If

PROOF : Take a rectangular axes paralle to the unit vector I,j,k and let were U,V,W are the components of F along the axes Now we have to prove that

kWjViUF

dxdydzs

z

W

y

V

x

U .ndSkWjViU

Where dx dy dz is the volume element dv. For fixed values of y,z take the rectangular Prism parallel to x-axis bounded by the planes y,y+dy,z,z+dz, the area of its normal Section being dydz. Such a prism cuts the boundary an even number of times at a point P1,P2,…..P2n, since the boundary surface is closed , if a point moves along the prism in The direction of x- increasing ,it enters the region at P1,p2,P3 and leaves it at the points P2,P4. ……..P2n

dz)dy U-.....UU(-U x

uILet 2n321

dxdydz

Where Ur is the value of U at point P Let dS be the area of the element of the boundary interceped by the prismat The point P Now dy dz is the area of projection of this element of the boundary intercepted

even isr if ,dSn.i

odd isr if dSn.i-dydz

rr

rr

Because the angle which the vector makes with is acute or obtuse according As r is given even or odd

)dSnU.....dSnUdSn(Ui I 2n2n2n222111

S

S

S

dSnkWdv

dSnjVdv

dSniUdv

ˆ.ˆz

W and

ˆ.ˆy

V Similarly,

ˆ.ˆx

U

V

V

V

dSn.F .dVFdiv

ˆ.kWjViU x

U Adding

V

SV

dSndvz

W

y

V

This theorem is also known as Divergence theorem

Cartesian equivalent of Divergence Theorem

n of cosinesdirection are cos,cos,cosaxes.Then ofdirection positivewith

makes n normalunit drawn outward which angles thebe ,,Let

FFFFdiv

kFjFiF FLet

321

321

zyx

dxdydzdxdydzdxdydzzyx

or

dSdxdydzzyx

321321

321321

321

FFFFFF

cosFcosFcosFFFF

as written becan Theorem Divergence

cosFcosFcosF n.F

k)(cosj)(cosi)(cos n

Questions 1. Prove that is the volume of the space enclosed by the surface S.

dSnr .3

1

Sol. By Gauss theorem

dSn.F .dVFdiv V

S surface by the enclosed space theof volume theis vwhere

.3

1

3v

3.

vdSnr

dvdivrdvndSr

dvWUdSnVUdvV

curlUandVcurlV

..

2

1

2

1

thatshow ,,2

1 W If

2

2.

dvWUdSnVUdv

dvWUdvdSnVU

dvWU

dvWUVV

dvVcurlUUcurlV

dvVUdivdSnVU

).(.2

1)V(

2

1

).()V(2

1 .

2

1

).2V(

)2..(

..

.

theoremsGauss' usingBy

2

2

Sol.

STOKE’s THEOREM : If f is any continuously differential vector point function and S is surface bounded by curve C then

is unit vector normal to surface

.

...

.

.

.

.

Green's theorem : It states that ,“if M(x,y) and N(x,y) be continuous functions of x and y having continuous partial derivatives ∂M and ∂N

∂y ∂x

in a region S of the x-y plane bounded by a

closed curve C , then

∮Mdx + Ndy =∬[∂N/∂x - ∂M/∂y] dxdy

C S

Where C is traversed in the counter clock wise

direction.

Proof : Consider an area in xy plane bounded by curve C . Let A be any differentiable vector field . Suppose M and N are x and y components of vector field A . Then A = Mî + Nĵ -----------(1)

The displacement vector dr in xy plane is

dr = dxî + dyĵ -----------(2)

Take the dot product of 1 and 2 . This gives

A . dr = (Mî + Nĵ ) . (dxî + dyĵ )

= Mdx + Ndy ------------(3)

z o y ds s x C

s

Now according to Stokes's theorem ∮A . dr = ∬(∇ ╳ A ) .dS C S

Now using eq. 3 , it becomes

∮Mdx + Ndy = ∬(∇ ╳ A ) . dS --------------(4) C S

Evaluation of ∇╳A and dS .

Let us evaluate

î ĵ k

∇ ╳ A = ∂ ∂ ∂

∂x ∂y ∂z

M N 0

SO ∇╳ A = - î ∂N + ĵ ∂M + k ∂N - ∂M ---(5)

∂z ∂z ∂x ∂y

the area element vector dS on the surface

S will point in z direction . Therefore

dS = dSk -------------(6)

using equation 5 and 6 in 4 , we get

(∇╳A ).dS = -î∂N +ĵ∂M +k∂N-∂M . (dSk)

∂z ∂z ∂x ∂y

so

(∇╳A).dS = ∂N-∂M .dS ----------(6)

∂x ∂y

Using equation 6 in equation 4 , we get

∮Mdx + Ndy =∬[∂N /∂x - ∂M /∂y] ds --(7)

C S

We know that dS is the area element in the xy

plane . Thus

dS = dxdy --------(8)

Using equation 8 in equation 7 , we get

∮Mdx + Ndy =∬[∂N/∂x - ∂M/∂y] dxdy

C S

This equation is known as the Green's theorem in

a plane .

Que. : Verify Green's theorem for ∫(xy+y2)dx +

x2dy , where C is the closed curve of the region bounded by y=x and y =x2.

Ans. : We know that by Green's theorem , we have

∮Mdx + Ndy =∬[∂N/∂x - ∂M/∂y] dxdy C S

Point of intersection : y = x , y = x2

i.e. x = x2

i.e. x(x-1) = 0

i.e. x = 0 or x = 1

Hence y = 0 or y = 1

Let M = xy+y2 and N = x2

∮Mdx+Ndy = ∮Mdx+Ndy+∮Mdx + Ndy

C c1 c2

= ∮{(xy+y2)dx+x2dy } +∮{(xy+y2)dx + x2dy c

1 c

2

=∫1{(x.x2+x4)dx+ydy}+∫

0{(x.x+x2)dx+y2dy}

0 1

= ∫1{(x3+x4)+ydy}+∫

0{2x2dx+y2dy}

0 1

= x4 + x5 + y2 1+ 2x3 + y3

0 4 5 2 0 3 3

1

= 1 + 1+ 1 + -2 + -1 4 5 2 3 3

= 19 – 1

20

= -1

20

Que. : Show that the area bounded by simple closed

curve is given by ½ ∫xdy-ydx and hence find the area of ellipse . c

Ans. :

½ ∫xdy-ydx = ½ ∬ ∂x /∂x -∂(-y) /∂y dxdy

c s

= ½ ∬ (1+1)dxdy s

= ∬dxdy s

= Area of surface S.

So area of ellipse = ½ ∫xdy-ydx c

= ½ ∫xdy-ydx -----------------(1) now normal equation of the ellipse is given by x2 + y2 = 1 a2

b2

now the parametric equation of the ellipse is given by x = acosθ , y = bsinθ -----------(2) using equation 2 in equation 1 , we get area of ellipse = ½ ∫2πacosθbcosθdθ + bsinθasinθdθ 0

= ½ ab ∫2π dθ 0

= ½ ab2π = πab .