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In a two-dimensional flow field given in cylindrical geometry with [tex]u_r = 0[/tex] and [tex]u_\theta = \frac{\gamma}{2\pi r}[/tex], show that:

a) [tex]\oint u \cdot dl = 0[/tex] if a closed loop [tex]l[/tex] does not enclose the origin.

Answer :

Final answer:

To show that the line integral of a vector field around a closed loop is zero, we can use Stokes' theorem. In this case, with the given velocity field, the curl of the velocity field is calculated to show that the surface integral of the curl is zero for surfaces that don't enclose the origin.

Explanation:

To show that ∮ u ⋅ dl = 0, we can use Stokes' theorem. Stokes' theorem relates the line integral of a vector field around a closed loop to the surface integral of the curl of the vector field over a surface enclosed by the loop.

In this case, since ur = 0 and uθ = γ/2πr, the curl of the velocity field is given by curl(u) = (1/r) ∂(ruθ)/∂r = (γ/2πr^2) eφ, where eφ is the unit vector in the azimuthal direction. Then, we can evaluate the surface integral of curl(u) over any surface enclosed by the closed loop, and by Stokes' theorem, the line integral of u around the loop is equal to this surface integral.

Since uθ is only dependent on the azimuthal component, the surface integral of curl(u) over any surface that doesn't enclose the origin will be zero. Therefore, we can conclude that ∮ u ⋅ dl = 0 if the closed loop l does not enclose the origin.

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