The picture shows a use of Ampere's law. A circular path is chosen.
$$\oint \vec B \bullet \mathrm{d}\vec l=\mu_0 I_{encl}$$
When using Ampere's law we are talking about the current enclosed. That means, the current through the surface enclosed by the path.
The text to the picture says:
For the plane circular area bounded by the circle, $I_{encl}$ is just the current $i_C$ in the left conductor. But the surface that bulges out to the right is bounded by the same circle, and the current through that surface is zero. So $\oint \vec B \bullet \mathrm{d} \vec l$ is equal to $\mu_0 I_{encl}$, and at the same time it is equal to zero! This is a clear contradiction.
(The topic is displacement current.)
I am confused of why they can simply let the area (the plane surface enclosed) bulge out like this. This means that the total current through any surfaces bounded by the same path must always be the same.
In the derivation earlier in the book of Ampere's law, only a plane surface was considered. What is the explanation that Ampere's law still applies when finding the line integral, nomatter the enclosed surface's shape in space?