Post on 17-Jan-2022
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
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
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:
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 ,,
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
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
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
.
..
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 ·
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)
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
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 .