I had to solve a problem where I had to calculate the work done by the force field given by: F → = ( − y , x ) x 2 + 4 y 2 , ( x , y ) ≠ ( 0 , 0 )

Question

I had to solve a problem where I had to calculate the work done by the force field given by:            F      →        =            (      −      y      ,      x      )                      x        2            +      4              y        2              ,  (  x  ,  y  )  ≠  (  0  ,  0  )

relientaaho2 · Accepted Answer

The force field             F      →        (  x  ,  y  )  =            (      −      y      ,      x      )                      x        2            +      4              y        2             is not conservative in   D  =            R        2    ∖  {  (  0  ,  0  )  }: by evaluating the line integral of             F      →       along the ellipse   γ,       x    2    +  4      y    2    =  4 counterclockwise, we find a non-zero result:      ∫          γ                  F      →        ⋅  d            s      →        =      ∫    0          2      π                  (      −      sin      ⁡      (      t      )      ,      2      cos      ⁡      (      t      )      )        4    ⋅  (  −  2  sin  ⁡  (  t  )  ,  cos  ⁡  (  t  )  )    d  t  =  πwhere   γ  (  t  )  =  (  2  cos  ⁡  (  t  )  ,  sin  ⁡  (  t  )  ) with   t  ∈  [  0  ,  2  π  ]In order to find the line integral along the unit circle       C    1   counterclockwise, we may apply the Green theorem to the domain   E  =  {  (  x  ,  y  )  :      x    2    +      y    2    ≥  1  ,      x    2    +  4      y    2    ≤  4  }      ∬    E        (                  ∂                  F          2                            ∂        x              −                  ∂                  F          1                            ∂        y              )      d  x  d  y  =      ∫          ∂      E                  F      →        ⋅  d            s      →      Since   ∂  E  =  γ  ∪      C    1    −   and             ∂              F        2                    ∂      x        −            ∂              F        1                    ∂      y        =  0, i.e.             F      →       is irrotational in E, it follows  0  =      ∬    E    0    d  x  d  y  =      ∫          γ                  F      →        ⋅  d            s      →        +      ∫                  C        1        −                        F      →        ⋅  d            s      →        =      ∫          γ                  F      →        ⋅  d            s      →        −      ∫                  C        1                        F      →        ⋅  d            s      →      and therefore      ∫                  C        1                        F      →        ⋅  d            s      →        =      ∫          γ                  F      →        ⋅  d            s      →        =  π  .

I had to solve a problem where I had to calculate the work done by the force field given by: <mr

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