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the folium of Descartes
the folium of Descartes is an algebraic curve defined by the
equation yxayx 333
When a = 1,
31
3
t
tx
3
2
1
3
t
ty
,0t
parametrically, we can
express part of the curve
of looped shaped as
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31
3
t
tx
3
2
1
3
t
ty
,0t
We have
31
3
t
tty
xty
.21
cdxydyxarea
.21
cdxxtxtdx
.21 2
cdxxtdxtxdtx
c
dtx221
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c
dtx221
31
3
td
cdt
t
t2
31
321
c
dtt
t23
2
1
921
dt
t
t23
2
1
33
so, area
c td
31
321
,0t
0 31
321
td
031
321
t
223
units
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MAT 128 1.0
iafgdalaiaf.a m%fushh
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iafgdalaiaf.a m%fushh Stokes’s Theorem
ds.Acurlrd.ASc
fuys mDIaG wkql,fha tall wNs,usnh i|yd Ok osYdj, c odrh osf.a
jdudj¾: osYdjg .uka lrk úg iqr;a moaO;shla idok osYdjg fjs.
S hkq ir, ixjD; odrhla iys; mDIaGhla hehs .ksuq. Let S be a surface and, c be the simple closed curved edge of S.
ika;;sl wjl, ix.=Kl iys; A ffoYsl lafIa;%hla i|yd For a vector field A, with continuous partial derivatives
fuys c hkq S mDIaGfha odrh fjs. c is the edge of the surface S.
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Stokes’s Theorem
S
c
dsAcurlrdASc
..
Here, positive direction for the surface integral is the unit vector making the right handed system when we move in anticlockwise direction along the curve. fuys mDIaG wkql,fha tall wNs,usnh i|yd Ok
osYdj, c odrh osf.a jdudj¾: osYdjg .uka lrk
úg iqr;a moaO;shla idok osYdjg fjs.
Eg: ffoYsl lafIa;%h iy
jyixA
0z,azyx 2222 mDIaGh ie,lSfuka
iafgdalaiaf.a m%fushh i;Hdmkh lrkak.
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Note: A line integral and a surface integral are connected by Stokes’s Theorem.
Eg: Verify the Stokes’s Theorem with and the surface .
jyixA 0z,azyx 2222
C : Where .
sinay,cosax 2,0
Acurl
.0sd.AcurlS
0
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c sd.A
djcosisina.jsinicosa2
0
2
0
2 dcossin2a
2
0
2 d2sina
d2cos21
a2
.0
Hence, Theorem is verified.
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jxiyB
0,2222 zazyx
sinay,cosax 2,0
Verify the Stokes’s Theorem with and the surface .
c sd.B
djcosisina.jcosisina2
0
2
0
2 da
.a2 2
E.g.
Bcurl k2 SS
sd.k2sd.Bcurl
S n.k
dAn.k2
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AdA2 Hence, Theorem is
verified.
2.2 a
E.g. Evaluate c
x dzdyy2dxe for
2z,4yx 22 The plane z = 2
c
x dzdyydxe 2
c
x dzkdyjdxi.kjyei 2
c
x rd.kjyei 2 Let kjyeiA x 2
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Acurl
1y2ezyx
kji
x
o
By Stokes theorem c
x dzdyydxe 2
c
sd.Acurl
c
sd.o
= o.
kjyeiA x 2
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c
dzysindyxcosdxzsin
along the boundaries of the rectangle3z,1y0,xo
E.g. Evaluate
kyjxizA sincossin
c
dzysindyxcosdxzsin
Choose
Acurl xkzjyi sincoscos
S
sd.Acurl
crd.A
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o π
1
S
dskxkzjyi .sincoscos
S
dsxsin
1sin
oyoxdydxx
.2
1cos oyox yx
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( 1, 0)( -1, 0)
( 0, 1)
( 0, -1)
c
dyxydxxy 2
c
dyjdxijxyixy 2
c
2 . rdjxyixy
E.g. Evaluate c
2 dydxxy xy
taken round the square C with vertices ( 1, 0),(-1, 0), ( 0, 1) and (0, -1)
S
sdyk .x-2
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( 1, 0)( -1, 0)
( 0, 1)
( 0, -1)
S
sdyk .x-2
S
dsy .x-2
1
1
20
1.x-
yx
yxydydxy
1
1
21
0.x-
yx
yxydydxy
dyyy
yx
yx
0
1
1
1
22 x21
-x dyyy
yx
yx
1
0
1
1
22 x21
-x
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dyyy
yx
yx
0
1
1
1
22 x21
-x dyyy
yx
yx
1
0
1
1
22 x21
-x
dyyyy
0
1
3 22 dyyyy 1
0
3 2x2
0
1
2421
yy
1
0
2421
yy
1
1
2421
yy
0
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Divergence Theorem
• Let S be a closed surface and V be the volume enclosed by S. Then outward flux of continuously differential Vector Field A over S is same as the volume integral of div A .
• i. e
VS
dvAcurlsd.A
wmidrs;d m%fushh
A hkq hus ixjD; mDIaGhla ;=< iy u; ika;;slj wjl,H ffoYsl flIa;
%hla kus, tu mDIaGh msrsjid msg;g A ys i%djh, mDIaGfhka wdjD;
jk m%foaYh msrsjid wmid A ys wkql,hg iudk fjs.
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The Divergence Theorem connects a volume integral and an integral over a closed surface. Letting.
,azyxS 2222
A( plane surface)
So ( curve surface)
0z
Note :
E.g.verify the Divergence Theorem for rA
ASS
sd.rsd.rsd.ro
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kzjyixa1
n
azyx
n.r222 a
az
n.k
For the curved surface So
n.kdA
n.rsd.rASo
az
dAa
A A
2
zdA
a
.a2 3 dAk.jyixAd.rAA
0
Hence, .a2sd.r 3
S
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Further, gives3rdiv VV
dv3dvAdiv
V
dv3
3a2
33
Hence, the Divergence Theorem is verified.
.a2sd.r 3
S
E.g.
Evaluate
S
2 sd.kzjyix , S is the closed surface bounded by the cone
222 zyx and the plane .1z
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Solution.
X
Y
Z
S
2 sd.kzjyix
V
2 dvkzjyixdiv
V
dvz22
VV
dvzdv 22
OGVV .22
OGV 12
43
11131
2 2 67
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S
22 dydxyxxyz2dxdzydzdyx*.1z,y,x0
Evaluate .
where S is the surface of the cube
S
3 sd.rr
* Suppose S is the surface of the sphere of radius a centered at O, evaluate
.
* For any closed surface S, evaluate S sd.r .
Here, Divergence Theorem can not be applied. Why?
ResultIf A.c = o for any constant arbitrary vector c , then A = 0 .
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1. If is a scalar field show that
vs
dvsd .
Solution : Let be an arbitrary constant vector.
Apply divergence Theorem to
c
ss
dvc.sd.c
c
ss
dv.csd.c
.dvsdss
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0sdrs
.
2. For any closed surface s, show that
Solution : Let be an arbitrary constant vector.c
ss
dsr.csdr.c
sd.rcs
dvrcdivv
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rcx
.ircdiv Since
,o gives us .0sdr.cs
rx
c.i
ii.c
Hence, . 0sdrs
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E.g. If s is the surface of the sphere 1zyx 222 Show that
.cba34
sd.kczjbyiaxs
Solution :
kczjbyiax. .cba
.dvcbasd.kczjbyiaxVs
.dvcbaV
.cba3
4
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S
22 dydxyxxyz2dxdzydzdyx*
.1z,y,x0
Evaluate .
where S is the surface of the cube
X
Y
Z
S
22 dydxyxxyz2dxdzydzdyx
S
kdydxjdxdzidzdy.A
kyxxyz2jyixA 22 Here
S
dsn.AApply divergence Theorem
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V
dvA. kyxxyz2jyixA 22
A. yxxy2y2x2
xy2
V
dvxy2
1
0z
1
0y
1
0x
dzdyydyx2
.2
1
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Evaluate E.g. .2 2222 c
dyyxdxyx
where c is the boundary of the surface in Oxy plane, enclosed
by Ox axis and the semi-circle .1 2xy
Flux form of Green’s Theorem.
Scdxdy
yN
xM
NdxMdy .%skaf.a m%fushfha i%dj wdldrh
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c
dyyxdxyx 22222
Scdxdy
yN
xM
NdxMdy
c
dxxydyyx 2222 2
2222 2,, xyyxNyxyxM
yxNy
yxMx
,, x2 y2
c
dyyxdxyx 22222 S
dsyx2
Now use polar coordinates.
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S
dsyx2
rdrdr
rsincos2
0
1
0
ddrr
rsincos2
0
21
0
0
1
0
3 cossin32
rr
2032
34
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jxiyyxA 2Verify the circulation form of the Green’s Theorem in Oxy plane for
and c is the curve and for
E.g.
.102 xxyxy
22 yx
jxiyF
Suppose that c is a simple closed curve Oxy plane not enclosing the origin O, evaluate
for
E.g.
rdFc
.