Post on 20-Oct-2019
UNIT-II
VECTOR CALCULUS
Directional derivative
The derivative of a point function (scalar or vector) in a particular
direction is called its directional derivative along the direction.
The directional derivative of a scalar point function φ in a given
direction is the rate of change of φ in the direction. It is given by the
component of gradφ in that direction.
The directional derivative of a scalar point function
φ (x,y,z) in the direction of →a is given by
→
→∇
a
a.φ.
Directional derivative of φ is maximum in the direction of φ∇ .
Hence the maximum directional derivative is φφ grador∇
Unit normal vector to the surface
If φ (x, y, z) be a scalar function, then φ (x, y, z) = c represents
A surface and the unit normal vector to the surface φ is given by
φφ
∇∇
Equation of the tangent plane and normal to the surface
Suppose →a is the position vector of the point ),,( 000 zyx
On the surface φ (x, y, z) = c. If →→→→
++= kzjyixr is the position vector of
any point (x,y,z) on the tangent plane to the surface at→a , then the
equation of the tangent plane to the surface φ at a given point →a on it is
given by 0. =
−→→
φgradar
If →r is the position vector of any point on the normal to the surface
at the point →a on it. The vector equation of the normal at a given point
→a on the surface φ is 0=×
−→→
φgradar
The Cartesian form of the normal at ),,( 000 zyx on the surface
φ (x,y,z) = c is
z
zz
y
yy
x
xx o
∂∂−=
∂∂−=
∂∂−
φφφ00
Divergence of a vector
If ),,( zyxF→
is a continuously differentiable vector point function in
a given region of space, then the divergences of →F is defined by
z
Fk
y
Fj
x
FiFdivF
∂∂+
∂∂+
∂∂==∇
→→
→→
→→→→
.
=x
Fi
∂∂
→→
∑
If →→→→
++= kFjFiFF 321 ,then ).( 321
→→→→++∇= kFjFiFFdiv
i.e., z
F
y
F
x
FFdiv
∂∂+
∂∂+
∂∂=
→321
Solenoidal Vector
A vector →F is said to be solenoidal if 0=
→Fdiv (ie) 0. =∇
→F
Curl of vector function
If ),,( zyxF→
is a differentiable vector point function defined at each
point (x, y, z), then the curl of →F is defined by
→→
×∇= FFcurl
= z
Fk
y
Fj
x
Fi
∂∂×+
∂∂×+
∂∂×
→→
→→
→→
= x
Fi
∂∂×
→→
∑
If →→→→
++= kFjFiFF 321 ,then )( 321
→→→→++×∇= kFjFiFFcurl
321 FFF
zyx
kji
Fcurl∂∂
∂∂
∂∂=
→
=
∂∂−
∂∂+
∂∂−
∂∂−
∂∂−
∂∂ →→→
y
F
x
Fk
z
F
x
Fj
z
F
y
Fi 121323
Curl→F is also said to be rotation
→F
Irrotational Vector
A vector →F is called irrotational if Curl 0=
→F
(ie) if 0=×∇→F
Scalar Potential
If →F is an irrotational vector, then there exists a scalar function φ
Such that φ∇=→F . Such a scalar function is called scalar potential of
→F
Properties of Gradient
1. If f and g are two scalar point function that ( ) gfgf ∇±∇=±∇ (or)
( ) gradggradfgfgrad ±=±
Solution: ( ) ( )gfz
ky
jx
igf ±
∂∂+
∂∂+
∂∂=±∇
→→→
= ( ) ( ) ( )
±∂∂+±
∂∂+±
∂∂ →→→
gfz
kgfy
jgfx
i
= z
gk
z
fk
y
gj
y
fj
x
gi
x
fi
∂∂±
∂∂+
∂∂±
∂∂+
∂∂±
∂∂ →→→→→→
=
∂∂+
∂∂+
∂∂±
∂∂+
∂∂+
∂∂ →→→→→→
z
gk
y
gj
x
gi
z
fk
y
fj
x
fi
= gf ∇±∇
2. If f and g are two scalar point functions then ( ) fggffg ∇+∇=∇ (or)
ggradffgradgfggrad +=)(
Solution: ( ) =∇ fg ( )fgz
ky
jx
i
∂∂+
∂∂+
∂∂ →→→
= ( ) ( ) ( )
∂∂+
∂∂+
∂∂ →→→
fgz
kfgy
jfgx
i
=
∂∂+
∂∂+
∂∂+
∂∂+
∂∂+
∂∂ →→→
z
fg
z
gfk
y
fg
y
gfj
x
fg
x
gfi
=
∂∂+
∂∂+
∂∂+
∂∂+
∂∂+
∂∂ →→→→→→
z
fk
y
fj
x
fig
z
gk
y
gj
x
gif
= fggf ∇+∇
3. If f and g are two scalar point function then 2g
gffg
g
f ∇−∇=
∇ where
0≠g
Solution: =
∇g
f
∂∂+
∂∂+
∂∂ →→→
g
f
zk
yj
xi
= ∑
∂∂→
g
f
xi
= ∑
∂∂−
∂∂
→
2g
x
gf
x
fg
i
=
∂∂−
∂∂
∑ ∑→→
x
gif
x
fig
g 21
= [ ]gffgg
∇−∇2
1
4. If →→→→
++= kzjyixr such that rr =→
,prove that →
−=∇ rnrr nn 2
Solution: nn r
zk
yj
xir
∂∂+
∂∂+
∂∂=∇
→→→
=
∂∂+
∂∂+
∂∂ →→→
z
rk
y
rj
x
ri
nnn
= z
rnrk
y
rnrj
x
rnri nnn
∂∂+
∂∂+
∂∂ −
→−
→−
→111
=
++→→→
−
r
zk
r
yj
r
xinr n 1
=
++→→→−
kzjyixr
nr n 1
= →−
rr
nr n 1
5. Find a unit normal to the surface 422 =+ xzyx at (2,-2, 3)
Solution: Given that xzyx 22 +=φ
)2( 2 xzyxz
ky
jx
i +
∂∂+
∂∂+
∂∂=∇
→→→φ
= ( ) ( ) ( )xkxjzxyi 222 2→→→
+++
At (2,-2, 3)
( ) )4()4(68→→→
+++−=∇ kjiφ
= →→→
++− kji 442
63616164 ==++=∇φ
Unit normal to the given surface at (2,-2,3)
φφ
∇∇
=6
442
→→→
++− kji
=
++−→→→kji 22
3
1
6. Find the directional derivative of xyzxzyzx ++= 22 4φ at (1,2,3) in the
direction of →→→
−+ kji2
Solution: Given xyzxzyzx ++= 22 4φ
)4( 22 xyzxzyzxz
ky
jx
i ++
∂∂+
∂∂+
∂∂=∇
→→→φ
= ( ) ( ) ( )→→→+++++++ kxyxzyxjxzzxiyzzxyz 842 222
At (1, 2, 3)
→→→
++=∇ kji 28654φ
Given: →→→→
−+= kjia 2
6114 =++=∴→a
→
→
∇=∴a
aDD .. φ
= 6
22.28654
→→→→→→ −+
++ kjikji
= [ ] [ ]866
1286108
6
1 =−+
7. Find the angle between the surface 5222 =++ zyx and
52222 =−++ xzyx at (0,1,2)
Solution: Let 222
1 zyx ++=φ and xzyx 2222
2 −++=φ
zz
yy
xx
2,2,2 111 =∂∂=
∂∂=
∂∂ φφφ
zz
yy
xx
2,2,22 222 =∂∂=
∂∂−=
∂∂ φφφ
→→→
++=∇ kzjyix 2221φ
→→→
++−=∇ kzjyix 22)22(2φ
At (o,1,2)
→→
+=∇ kj 421φ
→→→
++−=∇ kji 4222φ
Cos644416
422.42.
21
21
+++
++−
+=
∇∇∇∇=
→→→→→kjikj
φφφφθ
2420
20
2420
164cos =+=θ
= −
2420
20cos 1θ
=
−
24
20cos 1
8. Find the angle between the surfaces 1log 2 −= yzx and zyx −= 22 at the
point (1,1,1)
Solution: let zxy log2
1 −=φ and zyx += 2
2φ
z
x
zy
yz
x−=
∂∂=
∂∂−=
∂∂ 111 ,2,log
φφφ
1,,2 2222 =∂∂=
∂∂=
∂∂
zx
yxy
x
φφφ
→→→−+−=∇ kz
kjyiz 2)log(1φ
→→−=∇ kj22φ
Cos65
1
11414
12.
21
21 =+++
−=∇∇∇∇=
φφφφθ
= −
65
1cos 1θ
9. Find ( )nr2∇
Solution: ( )nr2∇ = ( )nr∇∇.
= ( ) ( ) ( )nnn rz
kry
jrx
i∂∂+
∂∂+
∂∂ →→→
= z
rnrk
y
rnrj
x
rnri nnn
∂∂+
∂∂+
∂∂ −
→−
→−
→111
→→→→
++= kzjyixr
222 zyxrr ++==
→
2222 zyxr ++=
r
x
x
rx
x
rr =
∂∂
⇒=∂∂
22
r
y
y
ry
y
rr =
∂∂
⇒=∂∂
22
r
z
z
rz
z
rr =
∂∂
⇒=∂∂
22
( )=∇∴ nr2
++→→→
−
r
zk
r
yj
r
xinr n 1
=
++→→→
− kzjyixnr n 2
=
→−
rnrn 2
Since →→→
+∇=
∇ udivuu φφφ .
( )
∇=∇→
− rnrr nn 22
= ( ) →−
→− ∇+
∇ rnrrnr nn .. 22
++
∂∂+
∂∂+
∂∂=∇
→→→→→→→kzjyix
zk
yj
xir.
=1+1+1 = 3
( ) ( ) →−− ∇+=∇ rrnnrr nnn .3 222
= ( )( ) 242 .23 rrnnnr nn −− −+
= ( )( )22 23 −− −+ nn rnnnr
( ) [ ] ( ) 2222 1 −− +=+=∇ nnn rnnnnrr
10. If →→→→
++= kzjyixr and rr =→
.Prove that →rr n is solenoidal if 3−=n and
→rr n is irrotational for all vectors of n.
Solution: →rr n
→→→++= krjyrixr nnn
div ( ) ( ) ( )zrz
yry
xrx
rr nnnn
∂∂+
∂∂+
∂∂=
→…………………(1)
Now 2222 zyxr ++=
Differentiating partially w.r.to x,
r
x
x
rx
x
rr =
∂∂
⇒=∂∂
22
Similarly, r
y
y
ry
y
rr =
∂∂
⇒=∂∂
22
r
z
z
rz
z
rr =
∂∂
⇒=∂∂
22
Now ( ) ( ) nnn rx
rr
rxxr
x+
∂∂
∂∂=
∂∂
.
= x.nnn r
r
xr +−1
( ) nnn rynryry
+=∂∂ − 22
( ) nnn rznrzrz
+=∂∂ − 22
From (1) we have
( ) nnn rzyxnrrrdiv 32222 +++=
−→
= nn rnr 3+
= ( ) nrn 3+
The vector →rr n is solenoidal if
→rrdiv n
= 0
( ) 03 =+⇒ nrn
03 =+⇒ n
3−=⇒ n →
∴ rr n is solenoidal only if n = -3
Now
zryrxr
zyx
kji
rrcurl
nnn
n
∂∂
∂∂
∂∂=
→→→
→
= ( ) ( )∑
∂∂−
∂∂→
yrz
zry
i nn
= ∑
∂∂−
∂∂ −−
→
z
rynrz
y
rnri nn 11
= ∑
− −−→
r
zynrz
r
ynri nn 11
= ( )∑ −−→
− yznryznri nn 22
= 0
Curl (→rr n ) =
→→→++ kji 000 =0
Curl (→rr n ) = 0 for all values of n
Hence →rr n is irrotational for all values of n.
11. Prove that ( ) ( )→→→→
+−++= kxzjxyizxyF 232 34sin2cos is irrotational and
find its scalar potential
Solution:
232 34sin2cos xzxyzxy
zyx
kji
Fcurl
−+∂∂
∂∂
∂∂=
→→→
→
= [ ] [ ] [ ] 0cos2cos23300 22 =−+−−−→→→
xyxykzzji →
∴F is irrotational.
To Find φ such that φgradF =→
( ) ( )z
ky
jx
ikxzjxyizxy∂∂+
∂∂+
∂∂=+−++∴
→→→→→→ φφφ222 34sin2cos
Integrating the equation partially w.r.to x,y,z respectively
),(sin 1
32 zyfxzxy ++=φ
),(4sin 2
2 zxfyxy +−=φ
),(33 yxfxz +=φ
,4sin 32 Cyxzxy +−+=∴φ is scalar potential
12. Prove that ).().(→→→→→→
−=
× BcurlAAcurlBBAdiv
Proof : ).(→→→→
×∇=
× BABAdiv
= ∑
×∂∂ →→→
BAx
i
= ∑∑
×
∂∂+
∂∂×
→→
→→
→→B
x
Ai
x
BAi
= ∑∑
×
∂∂+
×
∂∂−
→→
→→→
→B
x
AiA
x
Bi
= →
→→→
→→
∂∂×+
∂∂×− ∑∑ B
x
AiA
x
Bi ..
= →→→→
+
− BAcurlABcurl ..
13.Prove that →→→
∇−
•∇∇=
FFFcurlcurl 2
Solution:
×∇×∇=
→→FFcurlcurl
By using →→→→→→→→→
−
=
×× cbabcacba ..
= ( )→→
∇∇−∇
∇ FF ..
= →→
∇−∇
∇ FF 2.
VECTOR INTEGRATION
Line, surface and Volume Integrals
Problems based on line Integral
Example 1:
If ( ) →→→→+−+= kxzjyziyxF 22 201463 Evaluate ∫
→
C
drF . from (0,0,0) to
(1,1,1) along the curve 32 ,, tztytx ===
Solution: The end points are (0, 0, 0) and (1, 1, 1)
These points correspond to t = 0 and t = 1
23,2, tdztdydtdx ===∴
∫→
C
drF . = ( )∫ +−+C
dzxzyzdydxyx 22 201463
= ( ) ( ) ( )∫ +−+1
0
27522 32021463 dttttdttdttt
= ( )∫ +−1
0
962 60289 dtttt
= ( )101073 643 ttt +−
= ( )[ ] 50643 =−+−
Example 2:
Show that →→→→
++= kzjyixF 222 is a conservative vector field.
Solution: If →F is conservative then 0=×∇
→F
Now 0000
222
=++=∂∂
∂∂
∂∂=×∇
→→→
→→→
→kji
zyx
zyx
kji
F
→∴F is a conservative vector field.
Surface Integrals
Definition: Consider a surface S. Let n denote the unit outward normal to the
surface S. Let R be the projection of the surface x on the XY plane. Let→f be
a vector valued defined in some region containing the surface S. Then the
surface integral of →f is defined to be dydx
kn
nfdsnf
RS
.
.
.. ∫∫∫∫ →
→
∧→∧→
=
Example 1;
Evaluate dsnfS
∫∫∧→. where
→→→→−+= kzyjxizF 2
and S is the surface of
the cylinder 122 =+ yx included in the first octant between the planes z = 0
and z = 2.
Solution: Given →→→→
−+= kzyjxizF 2
122 −+= yxφ
→→
+=∇ jyix 22φ
22 44 yx +=∇φ
=222 yx +
=2
The unit normal ∧n to the surface =
φφ
∇∇
= yjxiyjxi +=+
2
22
xyxzjyixkzyjxiznF +=
+
−+=→→→→→∧→
.. 2
INTEGRAL THEOREMS
(i) Gauss’s divergence theorem
(ii) Stoke’s theorem
(iii) Green’s theorem in the plane
Green’s Theorem
Statement:
If M(x,y) and N(x,y) are continuous functions with continuous
partial derivatives in a region R of the xy plane bounded by a simple closed
curve C, then
dxdyy
M
x
NndyMdx
Rc
∫∫∫
∂∂−
∂∂=+ , where C is the curve described in the
positive direction.
Verify Green’s theorem in a plane for the integral ( ) xdydxyxc
+−∫ 2
taken around the circle 422 =+ yx
Solution: Green’s theorem gives
dxdyy
M
x
NNdyMdx
Rc
∫∫∫
∂∂−
∂∂=+
Consider ( ) xdydxyxc
+−∫ 2
M = x – 2y N = x
1,2 =∂∂−=
∂∂
x
N
y
M
dxdyy
M
x
N
R
∫∫
∂∂−
∂∂∴
( ) ∫∫∫∫ =+RR
dxdydxdy 321
= 3[Area of the circle]
= 32rπ
=3. 4.π
= π12 ……………………(1)
Now ∫ + NdyMdx
We know that the parametric equation of the circle 422 =+ yx
x = 2 cosθ y = 2 sinθ
θθddx sin2−= , θθddy cos2=
( ) xdydxxxNdyMdx +−=+∴ 2
= ( )( ) ( ) θθθθθθθ dd cos2cos2sin2sin4cos2 +−−
= θθθθθ d22 cos4sin8sincos2 ++−
Where θ various from 0 to π2
( )∫∫ ++−=+∴π
θθθθ2
0
2 4sin4sincos2 dNdyMdxC
= ∫
+
−+−π
θθθ2
0
42
2cos142sin d
= ( )∫ −+−π
θθθ2
0
2cos262sin d
=
πθθθ 2
02
2sin26
2
2cos
−+
= ππ 122
112
2
1 =−+ …………………….(2)
∴From (1) and (2)
dxdyy
M
x
NNdyMdx
Rc
∫∫∫
∂∂−
∂∂=+
Hence Green’s Theorem is verified.
Example 2
Using Green’s theorems find the area of a circle of radius r.
Solution: By Green’s theorem we know that
Area enclosed by C = ∫ −C
ydxxdy2
1
The parametric equation of a circle of radius r is x = θθ sin,cos ryr =
Where πθ 20 ≤≤
∴Area of the circle = ∫ −−π
θθθθθ2
0
)sin(sin)cos(cos2
1drrrr
= ( ) θθθπ
drr∫ +2
0
2222 sincos2
1
= ∫π
θ2
0
2
2
1dr
= [ ] 22
0
2
2
1rr πθ π =
Example 3:
Evaluate ( )[ ]∫ −−c
xdydxyx cossin where c is the triangle with
vertices (0,0) ,( )0,2
π and )1,
2(π
Solution: Equation of OB is
02
0
01
0
−
−=−−
πxy
πx
y2=⇒
By Green’s theorem dxdyy
M
x
NNdyMdx
Rc
∫∫∫
∂∂−
∂∂=+
Here 1,sin −=∂
∂−=y
MyxM
N xx
Nx sin,cos =
∂∂−=
( )[ ] ( )dxdyxxdydxyxRC
∫∫∫ +=−−∴ 1sincossin
In the region R, x varies from x = 22
ππtoy
and y varies from y = 0 to y = 1
( ) =−−∴ ∫ xdydxyxC
cossin ( )∫ ∫ +1
0
2
2
1sin dxdyxy
π
π
= [ ]∫ +−1
0
2
2
cos
π
πyxx dy
= dyyy
∫
−+1
0222
cosπππ
=
1
0
2
422sin2
−+ yy
y ππππ
= 2
2
42
2 ππ
πππ
+=−+
Example 4
Verify Green’s theorem in the plane for
( ) ( )∫ −+−C
dyxyydxyx 6483 22 where C is the boundary of the region defined
by
X = 0 , y= 0, x + y =1
Solution: We have to prove that
dxdyy
M
x
NNdyMdx
Rc
∫∫∫
∂∂−
∂∂=+
M = xyyNyx 64,83 22 −=−
yx
Ny
y
M6,16 −=
∂∂−=
∂∂
By Green’s theorem in the plane
dxdyy
M
x
NNdyMdx
Rc
∫∫∫
∂∂−
∂∂=+
= ( )∫ ∫−1
0
1
0
10 dydxy
x
= ∫−
1
0
1
0
2
210
x
y
= ( )∫ −1
0
215 dxx
= ( )
3
5
3
15
1
0
3
=
−− x
Consider ∫∫∫∫ ++=+BOABOAc
NdyMdx
Along OA, y=0 , x varies from 0 to 1
[ ] 131
0
3
1
0
2 ===+∴ ∫∫ xdxxNdyMdxOA
Along AB, y = 1 - x dxdy −=⇒ and x varies from 1 to 0 .
( ) ( ) ( )[ ]dxxxxxxNdyMdxAB
∫∫ −+−−−−=+∴0
1
22 1614183
= ( ) ( ) 0
1
32
232
232
14
3
18
3
3
−+
−−−
−−− xx
xxx
= 3
82312
3
8 =+−−+
STOKE’S THEOREM
If S is an open surface bounded by a simple closed curve C and if a vector
function →F is continuous and has continuous partial derivatives in S and on
C, then ∫∫∫→→→→
=c
rdFdsnFcurl .. where →n is the unit vector normal to the
surface (ie) The surface integral of the normal component of →Fcurl is equal
to the integral of the tangential component of →F taken around C.
Example 1
Verify Stoke’s theorem for ( )→→→→
−−−= kzyjyziyxF 222 where S is the upper
half of the sphere 1222 =++ zyx and C is the circular boundary on z = 0
plane.
Solution: By Stoke’s theorem ∫∫∫→→→→
=sc
dsnFcurlrdF ..
( )→→→→
−−−= kzyjyziyxF 222
zyyzyx
zyx
kji
Fcurl
222 −−−∂∂
∂∂
∂∂=
→→→
→
= [ ] ( ) ( )→→→→
=++−−+− kkjyzyzi 100022
Here →→
= kn since C is the circular boundary on z = 0 plane
=∴ ∫∫S
area of the circle ∫∫=→→
S
dxdydsnFcurl .
= ππ =2)1( ……….(1)
ON z = 0, ∫∫∫→→→→
=sc
dsnFcurlrdF ..
On C, x = cos θθ sin, =y
θθθθ ddyddx cos,sin =−=
θ varies from 0 to π2
( )( ) θθθθπ
drdFc
sinsincos2.
2
0
−−=∴ ∫∫→→
= ( ) ∫∫ +−ππ
θθθθθ2
0
2
2
0
sinsincos2 dd
= ( ) ∫∫
−+−ππ
θθθθ2
0
2
02
2cos12sin dd
=
ππ θθθ 2
0
2
0 2
2sin
2
1
2
2cos
−+
−
= ππ =++−2
1
2
1 ……………(2)
∴From (1) and (2)
∫∫∫→→→→
=sc
dsnFcurlrdF ..
Hence stoke’s theorem is verified
Example 2
Verify stoke’s theorem for ( ) ( )→→→→
−+++−= kxzjyzizyF 42 where s is
the surface of the cube x = 0, x = 2, y = 0, y = 2, z = 0 and z = 2 above the xy –
plane.
Solution:
By Stoke’s theorem
∫∫∫→→→→
=sc
dsnFcurlrdF ..
Given ( ) ( )→→→→
−+++−= kxzjyzizyF 42
xzyzzy
zyx
kji
Fcurl
−++−∂∂
∂∂
∂∂=
→→→
→
42
= [ ] [ ] [ ]1010 −++−−−→→→kzjyi
= [ ]→→→
−−+− kzjiy 1
Hence Stoke’s theorem is verified.
Example 3:
Verify Stoke’s theorem for →→→→
++= kxjziyF where S is the upper half
surface of the sphere 1222 =++ zyx and C is its boundary.
Solution: By stoke’s theorem
∫∫∫→→→→
=sc
dsnFcurlrdF ..
Gauss Divergence theorem
Statement:
The surface integral of the normal component of a vector
function F over a closed surface S enclosing volume V is equal to the volume
integral of the divergence of F taken throughout the volume V ,
dvFdsnFVS
∫∫∫∫∫→∧→
∇= ..
Evaluate zdxdyxydzdxxdydzx 223 ++∫∫ over the surface bounded by z = 0 ,z
= h, 222 ayx =+
Solution:
16
3
22
1
4
3cos
2
0
4 ππθθ
π
==∫ d
2
3.
4adsnF
S
=∫∫→→