Question

Show that the divergence of curl of a vector is zero.

Answer 1

 Let `bar(F) = F_x bar(a)_x + F_y bar(a)_y + F_z bar(a)_z`

`\bar(F) = ∇ xx bar(F) = [[a_x , a_y ,a_x],[∂/(∂x) , ∂/(∂y) , ∂/(∂z)],[F_x ,F_y ,F_z]]`

    ` =bar(a)_x xx ((∂F_x)/(∂_y) - (∂F_x)/(∂_z)) - bar(a)_y ((∂F_x)/(∂_x) - (∂F_x)/(∂_z))` 

      `+ bar(a)_z ((∂F_x)/(∂_x) - (∂F_x)/(∂_y))`

`" Now div (curl "bar(F)) = ∇ .(∇ xx bar(F))`

                                     `= ∂/(∂_x) ((∂F_z)/(∂_y) - (∂F_y)/(∂_z)) - ∂/(∂_y) ((∂F_z)/(∂_x) - (∂F_y)/(∂_z))`

  `+ ∂/(∂_z) ((∂F_Y)/(∂_x) -(∂F_x)/(∂_y))`

 `= (∂^2F_x)/(∂_x∂_y) - (∂^2F_y)/(∂_x∂_z) - (∂^2F_z)/(∂_x∂_x)+(∂^2F_x)/(∂_y∂_z)`

` +(∂^2 F_y)/(∂_z∂_x) - (∂^2 F_x)/(∂_z "_+∂_y) = 0`