The problem defines:velocity in the local base { e r , e θ , e z }, the gradient and acceleration in the local curvilignear base e i ⊗ e j avec i , j ∈ { r , θ , z }In a problem's solution, this result is used to calculate the stress resultant σ n → = [ 0 μ w / H − p a t m ] R = ∫ 0 2 π + ∫ 0 2 π ∫ 0 R d p a t m ⋅ e → z r d r d θ = = ∫ 0 2 π ( e → θ ( θ ) d θ ∫ 0 R d μ ω H r d r − ( 2 π p atm ∫ 0 R d r d r ) e → z then it says ∫ 0 2 π ( e → θ ( θ ) d θ = 0 which I've been stuck for one hour looking where it came from ?

Question

The problem defines:velocity in the local base   {      e    r    ,      e    θ    ,      e    z    }, the gradient and acceleration in the local curvilignear base       e          i        ⊗      e          j       avec   i  ,  j  ∈  {  r  ,  θ  ,  z  }In a problem&#039;s solution, this result is used to calculate the stress resultant  σ            n      →        =      [                            0                                      μ          w                      /                    H                                      −                      p                          a              t              m                                            ]    R  =      ∫          0              2      π        +      ∫          0              2      π            ∫          0                      R                  d                          p          a      t      m        ⋅                    e        →                    z        r  d  r  d  θ  =  =      ∫          0              2      π        (                    e        →              θ    (  θ  )  d  θ      ∫          0                      R        d                        μ      ω        H    r  d  r  −      (    2    π          p              atm                    ∫              0                              R                      d                                r    d    r    )                      e        →                    z      then it says       ∫          0              2      π        (                    e        →              θ    (  θ  )  d  θ  =  0 which I&#039;ve been stuck for one hour looking where it came from ?

etapi9a · Accepted Answer

The integral in question is, in some sense, a sum of all the radially-pointing unit vectors pointing out from the same &quot;anchor point&quot;. Since they all have equal magnitudes, all possible directions are included in this &quot;sum&quot;, and all angles are given equal weight, these vectors &quot;cancel out&quot; one another.An explicit demonstration uses the fact that                              e                →                    θ        (  θ  )  =  −  sin  ⁡  θ                    i            ^        +  cos  ⁡  θ                    j            ^        ,where                     i            ^       and                     j            ^       are the unit vectors (anchored at the same point) pointing along the x- and y-axes, respectively.                              ∫                      0                                2            π                                                                              e                            →                                            θ                          (        θ        )                  d                θ                            =                           ∫                      0                                2            π                          (        −        sin        ⁡        θ                                            i                        ^                          +        cos        ⁡        θ                                            j                        ^                          )                  d                θ                                          =                 −                                            i                        ^                                    ∫                      0                                2            π                          sin        ⁡        θ                  d                θ        +                                            j                        ^                                    ∫                      0                                2            π                          cos        ⁡        θ                  d                θ                                          =                 −                                            i                        ^                          ×        0        +                                            j                        ^                          ×        0                                          =                           0                ,            the zero vector.

The problem defines: velocity in the local base {e_r,e_theta,e_z}, the gradient and acceleration in the local curvilignear base e_i ox e_j avec i,j in {r,theta,z}

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