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

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

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

MAT 128 1.0

iafgdalaiaf.a m%fushh

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.

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.

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

c sd.A

djcosisina.jsinicosa2

0

2

0

2 dcossin2a

2

0

2 d2sina

d2cos21

a2

.0

Hence, Theorem is verified.

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

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

Acurl

1y2ezyx

kji

x

o

By Stokes theorem c

x dzdyydxe 2

c

sd.Acurl

c

sd.o

= o.

kjyeiA x 2

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

o π

1

S

dskxkzjyi .sincoscos

S

dsxsin

1sin

oyoxdydxx

.2

1cos oyox yx

( 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

( 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

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

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.

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

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

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

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

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 .

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

0sdrs

.

2. For any closed surface s, show that

Solution : Let be an arbitrary constant vector.c

ss

dsr.csdr.c

sd.rcs

dvrcdivv

rcx

.ircdiv Since

,o gives us .0sdr.cs

rx

c.i

ii.c

Hence, . 0sdrs

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

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

V

dvA. kyxxyz2jyixA 22

A. yxxy2y2x2

xy2

V

dvxy2

1

0z

1

0y

1

0x

dzdyydyx2

.2

1

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

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.

S

dsyx2

rdrdr

rsincos2

0

1

0

ddrr

rsincos2

0

21

0

0

1

0

3 cossin32

rr

2032

34

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

.