We have a wire going around in a helix (just like an inductor) and a constant magnetic field exists along its axis throughout the space.
How do we calculate the flux through it? I can't understand where are the area elements upon which to calculate $\int B \cdot da$.
In Figure-01 we have a solenoid of 1 turn. In case of uniform magnetic field $\mathbf{B}$ parallel to the axis of the helical wire the flux through the helical surface is equal to the flux through its projection on the $xy-$plane that is the circle shown. This equality, which is exact and not approximate, is proved in the Figure-03 below.
In Figure-02 we have a solenoid of 4 turns.
\begin{equation} \left. \begin{cases} \mathrm d\Phi_{\mathrm S}=\mathbf{B}\boldsymbol{\cdot}\mathrm d \mathbf{S}=\mathrm B\mathrm{ d S}\cos\theta \\ \mathrm d\Phi_{\Sigma}=\mathbf{B}\boldsymbol{\cdot}\mathrm d \boldsymbol{\Sigma}=\mathrm B\mathrm d \Sigma \end{cases}\right\}\quad \stackrel{\mathrm d \Sigma=\mathrm{ d S}\cos\theta }{=\!=\!=\!=\!=\!=\!\Longrightarrow\:}\mathrm d\Phi_{\mathrm S}=\mathrm d\Phi_{\Sigma} \tag{01}\label{01} \end{equation}
Figure-01.a was added a posteriori in order for the boundary of the helical wire of Figure-01 to be more clear.
I assume that you'd have no difficulty working out the flux linked with a closed circular loop if the flux density is uniform and normal to the plane containing the loop.
With a long solenoid (a helix of small pitch) of $n$ turns we usually treat each turn as a circular loop having, in the case you present, a flux $\Phi=\vec B. \vec A =BA$ through it. The flux linkage with the complete solenoid is then taken to be $nBA$.
Treating a helix as a series of closed loops may be thought unsatisfactory. I'm attracted to a bit of topological cleverness that regards the helix not as having $n$ disc-like surfaces, but as a single surface with $n$ folds, each of area very nearly equal to the disc area. Hard to describe and even harder to draw, I'm afraid.
