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∮C F⃗ . d ⃗r

41. Evaluate

from t = 0 to t = 2π.

where

⃗ F

= (x - 3y) i + (y - 2x) j and C is the closed curve in the xy plane, x = 2 cos t, y = 3 sin t

( Ans. 6π, if C is traversed in the positive (counterclockwise) direction.)

x = 2 cos t, y = 3 sin t are parametric equations of an ellipse with semi major axis a = 3 and semi minor axis b = 2.Ellipse is a simple closed curve and we traverse it in counterclockwise direction. ( Remember : Counterclockwise direction is taken positive and clockwise direction is taken negative.This is the convention.) For circle and ellipse parameter t is angle measured from positive x-axis in counterclockwise direction.

∮C F⃗ . d ⃗r =∮C [( x−3y) ̂i + ( y−2x) ̂j]. (dx ̂i + dy ̂j) put

x = 2 cost dx = -2sint dt

∮C ( x−3y )dx+ ( y−2x) dy

=

and y = 3sint When you traverse ellipse once , t changes from t = 0 to t = 2π dy = 3costdt

(t=2 π)

∮C F⃗ . d ⃗r =∫(t=0) [(2cost−3(3sint))(−2sint dt )+ (3sint−2( 2cost))3cost dt] (2 π) (2 π) 2 2 2 2 = ∫0 [−4sint cost+ 18sin t+ 9sintcost−12cos t]dt = ∫0 [5sint cost + 18sin t−12cos t ]dt (2 π) (1−cos 2t) (1+ cos2t ) −12 ]dt = ∫ [5sint cost + 18 0 2 2 =

(2 π)

∫0

[5sint cost + 9(1−cos 2t)−6(1+ cos2t)]dt =

(2 π)

∫0

[5sint cost + 9−9cos 2t−6+ 6cos2t]dt

(2 π) (sin 2 t ) (sin 2t) −3 + 3t] 2 2 0 5 3 5 3 2 = [ (sin (2 π))− (sin 2 (2 π))+ 3(2 π)]−[0] = [ (0)− (0)+ 6 π] = 6 π 2 2 2 2 2 ⃗ . d ⃗r around the triangle C of Figure 1, (a) in the indicated direction, (b) 43. If ⃗ F =(2x + y ) ̂i+ (3y−4x) ̂j , evaluate ∮C F

=

(2 π)

∫0

[5sint cost−3cos 2t + 3] dt = [5

Ans. (a) - 14/3 (b) 14/3

opposite to the indicated direction.

The problem is two dimentional as there is no k-component in Force field.

∮C F⃗ . d ⃗r =∮C [( 2x+ y 2) ̂i + (3y−4x) ̂j ] .(dx ̂i + dy ̂j)

=

∮C (2x+ y 2 )dx+ (3y−4x)dy

----------------> ( 1 )

FROM ( 0 , 0 ) TO ( 2 , 0 ) : x changes from x = 0 to x = 2 , y = 0 => dy = 0 then ( 1 ) becomes

∮C F⃗ . d ⃗r =∫C (2x+ y 2)dx+ (3y−4x) dy

=

(x=2)

∫(x=0) (2x+ ( 0)2) dx+ (3(0)−4x)(0)

FROM ( 2 , 0 ) TO ( 2 , 1 ) : y changes from y = 0 to y = 1 , x = 2 => dx = 0 , put in (1)

=

2

∫0 2x dx=[ x 2 ]20

=4

∮C F⃗ . d ⃗r =∫C ( 2x+ y 2) dx+ (3y−4x)dy =

1

∫0 (3y−8)dy =[

=

(y=1)

∫(y=0) (2 (2)+ y 2 )(0)+ (3y−4 (2)) dy

1 (3−16) −13 3 3y2 = −8y ] = [ −8]−[0 ]= 2 2 2 2 0

FROM ( 2 , 1 ) TO ( 0 , 0 ) : The equation of line joining (2 , 1 ) to ( 0, 0 ) is

( y− y 1) ( x−x 1) ( y−1) (x−2) ( y−1) ( x−2) = => => = => −2( y−1)=−1( x−2) = (0−1) (0−2) −1 −2 ( y 2−y 1 ) ( x 2−x1 ) 1 => −2y+ 2=−x+ 2 => y= x 2 1 1 So parametric equations are put x = t then y= t so dx = dt and dy= dt .For ( 2 , 1) t = 2 , For (0 , 0) t =0 i.e t changes 2 2

from t = 2 to t= 0.Now Evaluate ( 1 )

∮C F⃗ . d ⃗r =∫C ( 2x+ y 2) dx+ (3y−4x) dy

=

2

∫(t =2) (2t + ( 12 t) )dt+ (3( 12 t )−4t ) 12 dt (t =0)

0 1 2 1 2 3 3 1 t3 3 t 2 0 1 23 3 2 t + t−2t )dt = ∫2 ( t + t )dt = [ + ] = [0]−[ + 2 ] 4 4 4 4 4 3 4 2 2 4 3 8 −[4+ 9 ] −13 18 3 2 3 + ] = −[ + ] = = −[ = 43 2 3 2 6 6 0

= ∫2 (2t+

So the required integral along the closed curve in the indicated direction is

13

13

∮C F⃗ . d ⃗r =4− 2 − 6

=

[ 24−39−13] −28 −14 = = 6 6 3

14 3 ⃗ A=( x− y) ̂i+ (x+ y) ̂j .

In the opposite direction only the sign will change so the answer would be 44. Evaluate

∮C ⃗A . d ⃗r

around the closed curve C of Fig. 2

∮C ⃗A . d ⃗r =∮C [( x−y ) ̂i + ( x+ y ) ̂j].(dx ̂i + dy ̂j)

=

(Ans. 2/3)

∮C ( x−y )dx+ ( x+ y)dy

------------> ( 1 )

2

Along the curve y=x from ( 0 , 0 ) to ( 1 , 1 ) : Parametric equations of this curve are put x = t then y=t 2 For ( 0 , 0 ) t = 0 and for ( 1 , 1 ) t = 1 .Now Evaluate (1)

dy=2t dt

, dx = dt and

(t =1)

∮C ⃗A . d ⃗r =∫C ( x− y)dx+ ( x+ y )dy = ∫(t =0) (t−t 2) dt+ ( t+ t 2) 2t dt 1 2t 4 t 3 t 2 1 t 4 t3 t 2 1 3 2 = ∫0 (2t + t + t)dt = [ + + ] = [ + + ] = 4 3 2 2 3 2 0

Along the curve

0

=

1

∫0 (t−t 2+ 2t 2+ 2t3) dt

1 1 1 4 [ + + ]−[0] = 2 3 2 3

2

y =x from (1 , 1 ) to ( 0 , 0 ) :

Parametric equations of this curve are put x = t then

y 2=t =>

For ( 1 , 1 ) t = 1 and for ( 0 , 0 ) t = 0 .Now Evaluate (1)

∮C ⃗A . d ⃗r =∫C ( x− y)dx+ ( x+ y )dy

=

(t =0)

dy=

y=√ t , dx = dt and 1

∫(t =1) (t−√ t) dt+ (t+ √ t) (2 √ t) dt

=

0

1 dt ( 2 √ t) 1

1

∫1 (t−√ t+ 2 √ t+ 2 ) dt

1 ( + 1)

0 t2 1 1 t 2 1 1 1 1 1 −2 t 2 1 1 2 (3 /2) 0 = ∫1 (t+ − √ t)dt = [ + t− t ] = [0]−[ + − ] = ] = [ + t− 2 2 2 2 2 (3/2) 1 2 2 3 3 2 2 23 1 (4−2) Therefore the required integral along the given closed curve is ∮C ⃗A . d ⃗r = 43 − 23 = 3 = 23 0

45. If ⃗ A = (y - 2x) i + (3x + 2y) j , compute the circulation of A about a circle C in the xy plane with center at the origin and radius 2, if C is traversed in the positive direction. ( Ans. 8π ) ⃗A . d ⃗r ⃗ about a circle C is the line integral of ⃗ The circulation of A A around the circle that is we need to evaluate

∮C

along the circle in counterclockwise direction.

∮C ⃗A . d ⃗r =∮C [( y−2x) ̂i + (3x+ 2y ) ̂j] .(dx ̂i + dy ̂j)

=

∮C ( y−2x ) dx+ (3x+ 2y)dy

Parametric equations of a circle are x = r cost and y = r sint where r is the radius so x = 2 cost and y = 2 sint then dx = -2 sint and dy = 2cost .Once around a circle in counterclockwise direction t changes from t=0 to t= 2π.So

∮ ⃗A . d vecr =∮C ( y−2x )dx+ (3x+ 2y) dy (t =2 π) = ∫(t =0) (2sint−2 (2cost))(−2sintdt)+ (3(2cost)+ 2 (2sint )) 2cost dt (2 π) (2 π) 2 2 2 2 = ∫0 (−4sin t + 8sintcost+ 12cos t+ 8sintcost) dt = ∫0 (−4sin t+ 16sintcost+ 12cos t)dt (2 π) (1−cos2t ) (1+ cos2t ) = ∫ [−4 + 16sintcost+ 12 ] dt 0 2 2 = =

(2 π)

∫0 [−2 (1−cos2t)+ 16sintcost+ 6(1+ cos2t)]dt (2 π) (2 π) ∫0 [−2+ 2cos2t+ 16sintcost+ 6+ 6cos2t ] dt = ∫0 [4+ 8cos2t+ 16sintcost]dt

(sin2t ) (sin2 t) (2 π) 2 (2 π) + 16 ] = [4t+ 4sin2t+ 8sin t]0 2 2 0 2 = [4(2 π)+ 4sin2(2 π)+ 8sin ( 2 π)]−[ 0] = [8 π+ 0+ 0]=8 π 2 2 ϕ=2xy z+ x y , evaluate ∫C ϕdr where C = [4t+ 8

55. If

2

3

from t=0 to t=1 (b) consists of the straight lines from (0,0,0) to (1,0,0), then to (1,1,0), and then to (1,1,1). 19 ̂ 11 ̂ 75 ̂ 1̂ ( Ans. (a) i+ j+ k (b) j + 2 k̂ ) 45 15 77 2 (a) is the curve x = t,

(a)

y=t

,

z=t

1

∫C ϕdr=∫C (2xy2 z + x2 y)( dx ̂i + dy ̂j+ dz k̂ ) = ∫0 ( 2t (t 2)2 t 3+ t 2 (t 2))(dt ̂i + d (t 2) ̂j+ d (t 3) k̂ ) 1 1 8 4 2 8 4 8 4 8 4 2 = ∫0 ( 2t + t )( dt ̂i+ 2t dt ̂j+ 3 t dt k̂ ) = ∫0 [( 2t + t ) ̂i+ ( 2t + t ) 2t ̂j+ (2t + t )3 t k̂ ]dt 9 5 10 6 11 7 1 1 2t t 4t t 6t t 8 4 9 5 10 6 + 2 ) ̂j+ ( + 3 ) k̂ ] = ∫0 [( 2t + t ) ̂i + ( 4t + 2 t ) ̂j + (6t + 3t ) k̂ ]dt = [( + ) ̂i + ( 9 5 10 6 11 7 0 =

9

[(

5

1

2 1 2 1 6 3 2t t ̂ 2 10 1 6 ̂ 6 3 + ) i + ( t + t ) j + ( t11 + t7 ) k̂ ] = [( + ) ̂i+ ( + ) ̂j+ ( + ) k̂ ]−[0] 9 5 5 3 11 7 9 5 5 3 11 7 0

(10+ 9) ̂ (6+ 5) ̂ ( 42+ 33) ̂ 19 ̂ 11 ̂ 75 ̂ i+ j+ k = i+ j+ k 45 15 77 45 15 77

= (b)

FROM ( 0 , 0 , 0 ) TO ( 1 , 0 , 0 ) : x changes from x = 0 to x = 1 , y =0 => dy = 0 , z = 0 => dz = 0 So (x=1) ∫C ϕdr=∫C (2xy2 z + x2 y)( dx ̂i + dy ̂j+ dz k̂ ) = ∫(x=0) (2x (0)2 (0)+ x 2 (0))( dx ̂i + (0) ̂j+ (0)k̂ ) =

1

∫0 (0 ) dx ̂i

=0

FROM ( 1 , 0 , 0 ) TO ( 1 , 1 , 0 ) : y changes from y = 0 to y = 1 , x =1 => dx = 0 , z = 0 => dz = 0 So

∫C ϕdr=∫C (2xy2 z + x2 y)( dx ̂i + dy ̂j+ dz k̂ ) =

=

( y=1)

∫( y=0 ) ( 2(1) y2 (0)+ (1)2 y)((0) ̂i+ dy ̂j + (0) k̂ ) =

∫0 ( y) dy ̂j = [

2 1

y ] j 2 0

1̂ j 2

FROM ( 1 , 1 , 0 ) TO ( 1 , 1 , 1 ) : z changes from z = 0 to z = 1 , x =1 => dx = 0 , y = 1 => dy = 0 So (z =1 ) ∫C ϕdr=∫C (2xy2 z + x2 y)( dx ̂i + dy ̂j+ dz k̂ ) = ∫(z =0 ) ( 2( 1)(1)2 z + (1)2( 1))((0) ̂i+ (0) ̂j+ dz k̂ ) = = 2 k̂ Therefore along the whole curve C

56. If

1

⃗ F = 2y i - z j + x k , evaluate

( t=π/ 2 )

∫C ⃗F× d ⃗r =∫( t=0) =

(π/ 2)

∫0

y = sint dy = cost dt

= [z 2+ z]10 k̂

∫C ϕdr= 12 ̂j+ 2 k̂

∫C ⃗F× d ⃗r

along the curve x = cost, y = sin t, z = 2 cos t from t = 0 to t = π/2.

̂ ̂ ̂i + dy ̂j+ dz k) ∫C ⃗F× d ⃗r =∫C (2y ̂i −z ̂j+ x k)×(dx

put x = cost dx= -sint dt

1

∫0 ( 2z+ 1) dz k̂

( Ans.

1 (2− π ) ̂i+ (π− ) ̂j ) 4 2

z = 2cost dz = -2sint dt

( 2sint ̂i −2cost ̂j+ cost k̂ )×(−sint dt ̂i+ cost dt ̂j + (−2sint dt ) k̂ ) ̂j ∣ ̂i ∣ k̂ ∣ 2sint −2cost

cost ∣

∣−sint dt cost dt −2sint dt∣ (π/2)

̂i [4sint cost dt−cos2 t dt ]−̂j[−4sin 2 t dt + sintcost dt ]+ k̂ [2sintcost dt−2sintcost dt]

=

∫0

=

∫0 ̂i [4sint cost dt−cos2 t dt ]−̂j [−4sin 2 t dt + sintcost dt ] (π/2) (1+ cos2t) ( 1−cos2t) ∫0 ̂i [4sint cost dt− 2 dt ]− ̂j [−4 2 dt+ sintcost dt ]

=

(π/2)

2 (π/ 2) 2 (π/2) (sin t ) 1 1 ( sin2t ) ( sin2t) (sin t ) = ̂i [4 − t− ] − ̂j [−2t + 2 + ]

2

2 2 2 2 2 0 0 1 1 1 1 1 1 = ̂i [ 2 (sin2 π )− π − ( sin2 π )]− ̂j [−2 π + (sin2 π )+ (sin2 π )] = ̂i [ 2− π − (0 )]− ̂j [−π+ 0+ ] = (2− π ) ̂i+ (π− ) ̂j 2 22 4 2 2 2 2 2 4 4 2 4 2