30. Line integrals If + R F = P + Q k, is a vector field on - PDF document

30. Line integrals If  + R ˆ � F = P ˆ ı + Q ˆ k, is a vector field on R 3 and C is a curve in space then we can define the line integral � � F · d � r, C in the same way as we did in the plane. If we pick a parametrisation of C , r ( t ) = � x ( t ) , y ( t ) , z ( t ) � , and a ≤ t ≤ b � then we can express � d � r = � d x, d y, d z � = � v ( t ) d t and F = � P, Q, R � in terms of t and integrate this. If � F represents force, the line integral represents the work done moving a particle from the start P 0 to the end P 1 . Example 30.1. Let C be the parametric curve � r ( t ) = � t, t 2 , t 3 � � 0 ≤ t ≤ 1 and F = � yz, xz, xy � . C is called the twisted cubic. We express everything in terms of t , � r = � 1 , 2 t, 3 t 2 � d t F = � t 5 , t 4 , t 3 � . d � and So r = � t 5 , t 4 , t 3 � · � 1 , 2 t, 3 t 2 � d t = 6 t 5 d t. � F · d � The work done is then � 1 � 1 � � 6 t 5 d t = � t 6 F · d � r = = 1 . C 0 0 Now let’s suppose we go from P 0 = (0 , 0 , 0) to P 1 = (1 , 1 , 1) along a different path. Let’s say we go parallel to ˆ ı , C 1 , parallel to ˆ  , C 2 and then parallel to ˆ k , C 3 . Let C ′ = C 1 + C 2 + C 3 . So � � � � � � � � F · d � r = F · d � r + F · d � r + F · d � r. C ′ C 1 C 2 C 3 Parametrise C 1 in the obvious way r ( t ) = � t, 0 , 0 � 0 ≤ t ≤ 1 . � 1

Then � d � r = � 1 , 0 , 0 � d t and F = � 0 , 0 , 0 � . So � � F · d � r = 0 . C 1 Parametrise C 2 � r ( t ) = � 1 , t, 0 � 0 ≤ t ≤ 1 . Then � d � r = � 0 , 1 , 0 � d t and F = � 0 , 0 , t � . We have � F · d � r = � 0 , 0 , 1 � · � 0 , 1 , 0 � d t = 0 d t. So � � F · d � r = 0 . C 2 Parametrise C 3 � r ( t ) = � 1 , 1 , t � 0 ≤ t ≤ 1 . Then � d � r = � 0 , 0 , 1 � d t and F = � t, t, 1 � . We have � F · d � r = � t, t, 1 � · � 0 , 0 , 1 � d t = d t. So � 1 � � F · d � r = d t = 1 . C 3 0 The reason why both answers are the same is that � F is a gradient vector field, � F = � yz, xz, xy � = ∇ ( xyz ) = ∇ f, where f ( x, y, z ) = xyz . As before we have the fundamental theorem of calculus for line integrals � ∇ f · d � r = f ( P 1 ) − f ( P 0 ) . C As before this means the integral is path independent and � F is conser- vative. In our case f (0 , 0 , 0) = 0 f (1 , 1 , 1) = 1 , so that f (1 , 1 , 1) − f (0 , 0 , 0) = 1 . F be a vector field on R 3 (or more generally a Theorem 30.2. Let � simply connected region; more about this later). � F is a gradient vector field if and only if P y = Q x , P z = R x and Q z = R y . 2

Again, one direction is reasonably clear. If � F = ∇ f, then P = f x , Q = f y and R = f z , so that P y = f xy and Q x = f yx , so that the equality P y = Q x is just saying the mixed partials of f are equal. Example 30.3. For which a and b is ı + ( x 2 + z 3 )ˆ  + ( byz 2 − 4 z 3 )ˆ � F = axy ˆ k a gradient vector field? We have Q = x 2 + z 3 R = byz 2 − 4 z 3 . P = axy, and Q x = 2 x and P y = ax , so that ax = P y = Q x = 2 x, and so a = 2. Q z = 3 z 2 and R y = bz 2 so that bz 2 = R y = Q z = 3 z 2 and so b = 3. Finally, P z = 0 and R x = 0 , so P z = R x is clear. Hence ı + ( x 2 + z 3 )ˆ  + (3 yz 2 − 4 z 3 )ˆ � F = 2 xy ˆ k is conservative. Let’s look for a potential function f ( x, y, z ). We want to solve three PDEs ∂f ∂f ∂f ∂y = x 2 + z 3 ∂z = 3 yz 2 − 4 z 3 . ∂x = 2 xy, and We have ∂f ∂x = 2 xy. Integrate both sides with respect to x . � 2 xy d x = x 2 y + g ( y, z ) . f ( x, y, z ) = 3

Note that the constant of integration is actually a function of both y and z . In other words ∂g ( y, z ) = 0 . ∂x Let’s take this answer for f and plug this into the second PDE. x 2 + ∂g ( y, z ) ∂f ∂y = x 2 + z 3 = x 2 + z 3 . so that ∂z Cancelling we get ∂g ( y, z ) = z 3 . ∂y Integrating this with respect to y we get � z 3 d y = z 3 y + h ( z ) , g ( y, z ) = where the constant of integration is an abritrary function of z . So now we know f ( x, y, z ) = x 2 y + z 3 y + h ( z ) . Finally, we plug this back into the third PDE: ∂f 3 yz 2 + dh ∂z = 3 yz 2 − 4 z 3 dz = 3 yz 2 − 4 z 3 . so that Cancelling we get dh dz = − 4 z 3 . Integrating with respect to z , we get h ( z ) = − z 4 + c. We can take c = 0. Putting all of this together f ( x, y, z ) = x 2 y + z 3 y − z 4 . One can define a vector field which measures how far � F is from being conservative. curl � F = ∇ × � F ˆ � � ˆ ˆ ı  k � � � ∂ ∂ ∂ � = � � ∂x ∂y ∂z � � P Q R � � � ∂ ∂ � � ∂ ∂ � � ∂ ∂ � � + ˆ � � � � � � = ˆ ı ∂y ∂z � − ˆ  ∂x ∂z k ∂x ∂y � � � � � � P R Q R P Q � � � �  + ( Q x − P y )ˆ = ( R y − Q z )ˆ ı − ( R x − P z )ˆ k. 4

This is the curl of the vector field � F . It measures the angular velocity. For example, � v = �− ωy, ωx, 0 � , represents rotation around z -axis with constant angular velocity ω . v = 2 ω ˆ curl � v = ∇ × � k. So the magnitude is twice the angular velocity and the direction is the axis of rotation. 5