Vector Analysis and Geometry

11.05.2012

E- Content in Mathematics Subject Expert: Bijumon R

Content Editor: Nandakumar

PRESENTER:RASHMI.M

10 STOKE’S THEOREM

Objectives

From this unit a learner is expected to achieve the following

1. Familiarize stoke’s theorem 2. Learn the definition of surface integral of a vector field. 3. Learn the method of finding surface integral of a vector field.

Sections

In this session we discuss Stokes’ theorem which says that the circulation of vector field F around the boundary C of an oriented surface S in the direction counterclockwise with respect to the surface’s unit normal vector n equals the integral of  F  n over S.

Surface Integrals We now show how to integrate a function over a surface.

1 Definitions If R is the shadow region of a surface S defined by the equation f( x , y , z )= c , and g is a continuous function defined at the points of S, then the integral of g over S is the integral f 蝌g( x , y , z ) dA , R 炎f p where p is a unit vector normal to R and f p  0 . The integral itself is called a surface integral. The Surface Area Differential and the Differential Form for Surface Integrals f Surface area differential is dS= dA and differential formula for surface integral is 炎f p

蝌g dS. Hence S f 蝌g dS= 蝌 g( x , y , z ) dA . S R 炎f p

We also recall the following surface integral formulae.

dxdy F.n dS  F.n ,    n.k where R is the orthogonal projection of S R

S on the XY plane.

dydz F.ndS  F.n    n.i , where R1 is the projections of S on YZ plan S R1

dzdx F.ndS = F.n  蝌蝌 n.j , where R2 is the projection of S on ZX plane. S R2

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Stokes’ Theorem Now we state Stokes’ Theorem.

2 Theorem (Stokes' theorem) The surface integral of the normal component of the curl of the vector field F, taken over a bounded surface S, equals the line integral of the tangential component of the vector F , taken over the closed curve C bounding the surface S.  F.n dS  F .dr i.e.   , S C where n is a unit vector normal to the surface, and F is a vector point function.

Example 1 Verify Stokes' theorem for F = (x2 - y2) i + 2xy j in the rectangular region in the xy plane, the vertices of the rectangle being given by (0, 0), (a, 0), (b, 0) and (a, b).

Solution

Here the bounded surface S is the rectangle R given by

R: 0＃ x a , 0 ＃ y b .

(Ref. Fig. 1)

Given F = (x2 - y2) i + 2xy j .

i j k       F   i 0  j 0  k 4y  4y k. x y z x2  y2 2xy 0

The surface is the rectangular region in the xy plane. Hence the outward normal to the surface is k.

 n dS = k dx dy.

a b a b \蝌(汛 F) .S = 蝌(4y k) .( k dy dx) 蝌 4 y dy dx = 2 ab2 . ____ (1) S 0 0 0 0

Now 蝌F.d r= F. d r + 蝌 F. d r + F. d r + F. d r ____ (2) C OA AC CB BO

(Ref figure)

(0, b) B C(a, b)

(0,0) O A (a, 0) X

3 Fig. 1

F.dr = [ (x2  y2) i + 2xy j ] . [ dx i + dy j + dz k ]

= (x2  y2)dx + 2xy dy

(i) In OA, y = 0 , dy = 0 and x varies from 0 to a.

a a 2 2 2 1 3 F.dr  ( x  0 )dx  2x.0.0  x dx  a .    3 OA 0 0

Similarly,

b ii  F.dr   2aydy  ab2 . ( x  a dx  0.) AC 0

0 a 1 iii F.dr  ( x2  b2 )dx   ( x2  b2 )dx  ab2  a3 ( y  b dy  0.)    3 CB a 0 iv  F.dr  0. ( x  0 dx  0.) BO

Using the above values, (2) becomes

1 1 F.d r = a3 + ab 2 + ab 2 - a 3 = 2 ab 2 3 3 ____ (3) C

From (1) and (3), we get

 F.n dS  F .dr   , S C so that Stokes’ theorem is verified.

Example 2 Verify Stokes' theorem for F = (2x – y) i  yz2 j  y2z k, where S is the upper half surface of the unit sphere x2 + y2 + z2 = 1 and C is its boundary.

Solution

i j k      F   k on simplification. x y z 2x  y  yz2  y2z

4 dx dy \蝌(汛F) � ndS 汛蝌( F) n , S R n k

where R is the orthogonal projection of S on the xy-plane. i.e., R is the region enclosed by the circle x2+ y 2 = 1.

dx dy = 蝌k n , R n k

= 蝌dx dy R

Now 蝌dxdy is the area of the region enclosed by the circle x2+ y 2 =1 and is p�12 p . Hence, R we obtain

汛蝌( F) .n dS = π, ____ (4) S

The boundary C of S is a circle in the xy plane of radius 1 and center at the origin. Suppose x = cos t, y = sin t, z = 0, 0  t  2 be the parametric form of C.

蜒蝌F. dr= (2 x - y ) dx (Q z = 0, dz = 0) C C

2  (2cost  sin t )(  sin t ) dt (since x  cos t , dx   sin tdt ) 0

=  (5)

From (4) and (5), we get

 F.n dS  F .dr   S C

Hence Stokes' theorem is verified.

Example 3 Use Stokes' theorem to evaluate curlF .n dS over the upper half of the hemisphere S of radius a, center at the origin, if

F = 2y i  x j + z k.

5 Solution By Stokes' theorem,

curl F.ndS   F.n dS  F . dr    , S S C where C is the boundary of S. Here boundary C of S is a circle in the xy plane of radius a and center at the origin. i.e. C is the circle x2+ y 2 = a 2 .

F.dr = (2y i  x j + z k) . ( dx i + dy j + dz k) = 2y dx  x dy + z dz.

Along C, z = 0, so that dz = 0. Hence

F.dr  2ydx  xdy    C C C

Suppose x = a cos  , y = a sin  are the parametric equations of the curve C.

2π 2π  F .dr   2a sinθ(  a sinθ)dθ  (a cosθ)(acosθ)dθ C 0 0

2  a 2  (2sin 2 θ  cos 2θ)dθ  3πa 2 . 0

Therefore, the required integral is given by

2 curlF.ndS  3πa . S

r.dr  0. Example 4 Prove that  C

Solution

By Stokes’ theorem

r.dr    r.ndS  0. ( curl r  0.)   C S

6 xy dx  xy 2dy Example 5 Evaluate  by Stoke’s theorem where C is the square in the xy- plane C with vertices (1, 0), (- 1,0), (0,1), (0, - 1).

Solution Here F.dr  xy dx  xy 2dy , so that F  xy i  xy 2 j and hence

i j k     2    F    y  xk. x y z   xy xy2 0

Also n = k.  curl F . n = (y2  x) k . k = y2  x.

 xy dx  xy 2dy  F.n dS  y 2  xdx dy    C S S

1 1 1  2 1 2  2 4  y x  x  dy  2y dy  . 1  2  1 1 3

Example 4 Verify Stokes Theorem. Given the hemisphere S: x2+ y 2 + z 2 = 9, z 0 , its bounding circle C: x2+ y 2 = 9, z = 0, and the field F=y i - x j . Solution We calculate the counterclockwise circulation around C (as viewed from above) using the parameterization (of the given circle) r( )  (3cos  ) i  (3sin  ) j , 0�� 2 : dr=( - 3sin q d q ) i + (3cos q d q ) j F=y i - x j =(3sin q ) i - (3cos q ) j F�d r - q9sin2 q d - q 9cos 2 q d = - q 9 d

2 Fd r  9 d    18  . C 0 P  N   M  P    N  M  Also,  F    i    j     k y  z   z  x    x  y  (0  0)i  (0  0) j  (  1  1) k   2 k xi y j  z k x i  y j  z k n   is the outer unit normal x2 y 2  z 2 3

With f( x , y , z )= x2 + y 2 + z 2 = 3 2 , z 0 , we take p = k and calculate

�f+2 xi + 2 y j = 2 z k + 2 x2 + y 2 = z 2 � 2 3 6 f p   f  k 2 z  2 z

f 3 dS= dA = dA 炎fp z

7 2z 3 汛 F� n dS - = dA - 2 dA 3 z Hence the curl integral of F is 汛 F� n dS - =2 dA - p 18 . 蝌蝌 . S x2+ y 2 9 Hence Stoke’s Theorem is verified.

Example 6 A fluid of constant density d rotates around the z-axis with velocity v ( y i  x j ) where w is a positive constant called the angular velocity of the rotation. If F  v, find   F and relate it to the circulation density. Solution With F v  y i   x j P  N   M  P    N  M   F    i    j     k y  z   z  x    x  y  =(0 - 0)i + (0 - 0) j + ( dw - ( -dw )) k = 2 dw k . By Stokes’s theorem, the circulation of F around a circle C of radius r bounding a disk S in a plane normal to   F, say the xy-plane, is F�d r汛 �� F d n dS w2 �d k w k dxdy p (2 r )(2 ) 蝌C 蝌 S S 1 Thus ( F )  k  2   F  d r . … (6) 2 C Eq. (6) relates 汛 F with the circulation density. F d r, F=xz i + xyj+ 3 xz k Example 7 Use Stokes’s theorem to evaluate C if and C is the boundary of the portion of the plane 2x+ y + z = 2 in the first octant traversed counterclockwise as viewed from above (Fig. 3). Solution The plane is the level surface f( x , y , z )= 2 of the function f( x , y , z )= 2 x + y + z .The unit normal vector f (2i  j  k ) 1 n  (2 i  j  k ) f 2i  j  k 6 is consistent with the counterclockwise motion around C. To apply stokes’s theorem, we find i j k 抖 curlF=汛 F = = (x - 3 z ) j + y k 抖x y z xz xy3 xz On the plane, z=2 - 2 x - y , so 汛 F=(x - 3(2 - 2 x - y )) j + y k = (7 x + 3 y - 6) j + y k 1 1 and  F  n (7x  3 y  6  y )  (7 x  4 y  6) 6 6 The surface area element is

8 f 6 d  dA  dxdy . f k 1 Using Stokes’s theorem, the circulation is Fd r    F n d  C  S 1 2 2x 1 (7x  4 y  6) 6 dy dx 0  0 6

1 2 2x (7x  4 y  6) dy dx   1. 0  0

Summary

In this session we discussed surface integrals and differential form of surface integrals and also stokes theorem

Assignments Verify Stokes’ theorem for the following data.

1. F = (x2 + y2) i  2xy j taken around the rectangle bounded by the lines x = a, y = 0 and y=b. 2. F = y i + 2x j + 2 k and C is the circle x2 + y2 = 1 in the xy plane and S is the plane area bounded by C. 3. F = y2 i + xy j  xz k on the upper half of the sphere x2 + y2 + z2 = a2. 4. F = x2 i + xy j in the square region in the xy-plane bounded by the lines x = 0, y =0, x=a, y=a. 5. F = y i + 2 j + x k where the bounding surface is the hemisphere x2 + y2 + z2 = 1, z  0. 6. F = (2y + z) i + (x  z) j + (y  x) k taken over the triangle ABC cut off from the plane x + y + z =1 by coordinate planes.

1. Given F = (x2 + y2 4) i + 3xy j + (2xz + z2) k and S is the surface of the hemisphere x2 + y2 + z2 =

F.ndS 16 above the xy plane. Then, using Stokes’ theorem, the value of  is ______. S

(a) 16 

(b) 16 

9 (c) 4 

(d) 4 

Ans. (a)

2. If C is the curve of intersection of x2 + y2 + z2  2ax  2ay = 0 with x + y = 2a, then the value of ydx  zdy  xdz    is ______. C C C (a) 2 2p a2 .

(b) -2 2p a2 .

(c) 2p a2 .

(d) 2p a2 .

Ans. (b)

1. What is circulation?

Ans. If C is a simple closed curve ( i.e., a curve which does not intersect itself anywhere), then the line integral of F around C is called the circulation of F about C and is denoted by Fd r . C

2. What is circulation density????

Ans. The circulation density or curl of a vector field F=M i + N j at the point (x , y ) is

N  M curlF   . x  y

Glossary

Stokes' theorem: The surface integral of the normal component of the curl of the vector field F, taken over a bounded surface S, equals the line integral of the tangential component of the vector F , taken over the closed curve C bounding the surface S.

10  F.n dS  F .dr i.e.   , S C where n is a unit vector normal to the surface, and F is a vector point function. surface integral:If R is the region of a surface S defined by f( x , y , z )= c , and g is a continuous function at the points of S, then the integral of g over S is the integral f 蝌g( x , y , z ) dA , R 炎f p f p  0 . The integral itself is called a surface integral.

REFERENCES Books

1. Murray R. Spiegel, Vector Analysis, Schaum Publishing Company, New York. 2. N. Saran and S. N. Nigam, Introduction to Vector Analysis, Pothishala Pvt. Ltd., Allahabad. 3. Erwin Kreyszig, Advanced Engineering Mathematics, John Wiley & Sons, 1999. 4. Shanti Narayan, A Text Book of Vector Calculus, S. Chand & Co., New Delhi.

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