Line Integrals

 

 

Example:   Evaluate the line integral for   P(x,y) = y2  and  Q(x,y) = x    along the curve, C,

given by   x  =  y3   from  (-1, -1) to (1, 1).

 

   I  =       ∫  y2  dx  +    ∫ xdy                   For the given curve, C,   dx  =  3y2 dy

                C                  C

                                                              1                         1

So the line integral becomes    I  =      ∫  y2  3y2 dy  +    ∫  y3 dy

                                                             -1                        -1

                                           1

I  =  (3/5)y5   +  (1/4) y4   |     =  [(3/5) +)] – [-(3/5) + (1/4)]  =   6/5

                                          -1

 

 

 

 

 

Example:     Evaluate the line integral   I =   ∫  f(x,y,z) ds  for      0 ≤  t  ≤  1     where

                                                                         C 

 

f(x,y,z)  =  2x + 9xy      along the curve, C,      given by  x  =  t,  y  =  t2,  z  =  t3

 

I =   ∫  f(x,y,z) ds  =   ∫  f(x,y,z) (ds/dt) dt  , ds/dt  =  √[ dx/dt)2 + (dy/dt)2 + dz/dt)2]dt

 

             1                                                                 1

     I =   ∫  [2t + 9 t t2 ]  √ [1 + (2t)2 + (3t2)2] dt  =  ∫  [2t + 9t3] √ [1 + 4t2 + 9t4] dt

            0                                                                 0

 

Let    w  =  1 + 4t2 + 9t4],  then  dw  =  4 (2t + 9t3) dt 

 

or   (2t + 9t3) dt   =  (1/4) dw   so the integral becomes

 

             14                                                     14

    I  =   ∫  (1/4) w1/2 dw  = (1/4)[(2/3)w3/2)] |      =  (1/6)[14√14 - 1]

            1                                                        1

 

 

 

 

 

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