What does the surface integral tell us, and how is it different from surface area? In the context of Faraday's Law of Induction,
Will a paraboloid with the same cross sectional area as a major portion of the sphere (Spherical Cap), have the same magnetic flux, when put in an uniform magnetic field ?
If it does have the same flux, does it mean that both the figures have the same surface integral ?
What does the surface integral tell about surfaces which do not lie in a single plane, that is not all points on the boundary of the surface are co-planar, and how is it different from surface area?
Will two surfaces having the same boundary have the same surface integral?
 A: In the equation $$\Phi=\int_{\mathbf{s}}\bf B\cdot dA$$ there is a dot product between the magnetic field and the normal area vector. This is evaluated at all points on the surface $\mathbf{s}$ and each one of these dot products depends on the angle between $\bf B$ and $\bf dA$.
For surfaces that are of different shape or orientation, the dot product will be different at points on each surface. So even if the total surface area between two surfaces is the same, the value of $\Phi$ will differ in general.
So a paraboloid and the section of a sphere will not have the same value for $\Phi$ even if the total surface areas and $\bf B$ (which includes the direction of $\bf B$ of course) are the same for each.
The same logic applies to the rest of your question. But:
Will two surfaces having the same boundary have the same surface integral?
Yes.
Any two surfaces that have the same boundary will have equal amounts of flux. For example, if you have a magnet inside a large sphere and slice off a section off the top of the sphere so that you have a very small hemisphere (so that it almost looks like a flat disc) and a larger one being the rest of the sphere, that both of these spheres have the same boundary will mean they have equal amounts of flux.
The number of lines of forces going through the larger hemisphere will be significantly higher than the smaller section, but the number of lines of force coming back into the larger section will cancel most of those leaving so that the net flux going through both surfaces turns out to be equal.
A: In general, a surface integral is exactly the same as an area integral, the only key difference is that it measures the surface area of a non flat surface.
For the same boundary, $\iint da$ will have a different value for every singe surface attached to that closed boundary, because the surface area of a surface depends upon the surface.
Now we have faradays law, which instead of a surface integral, it is a weighted sum of $\frac{\partial \vec{B}} {\partial t} \cdot \vec{da}$. This isn't just a measure of surface area, it is a measure of how much the time derivative of the magnetic field points directly into a chosen surface, * area.
There is a major difference between a regular surface integral and faradays law.
$\oint \vec{E} \cdot \vec{dl} = -\iint \frac{\partial \vec{B}} {\partial t} \cdot \vec{da}$
Here we have a line integral that only depends on the BOUNDARY, equated to a surface integral that shares a boundary with the line integral. This surface integral IS INDEPENDANT of the surface chosen, and unlike the regular surface integral, it only depends upon the boundary, not the surface.
Why is there a difference?
The key in understanding this is that for the integral to be independant of the surface, the inside function must have zero divergence.
Mathematically, we can easily prove this by assuming that this is true for all surfaces, and then finding the condition on B upon which this statement is true, and showing that it is indeed true.
$\oint \vec{E} \cdot \vec{dl} = -\iint\frac{\partial \vec{B}} {\partial t} \cdot \vec{da}$
$\iint \nabla × \vec{E} \cdot \vec{da} = -\iint\frac{\partial \vec{B}} {\partial t} \cdot \vec{da}$
The fact that I cross the surface off here indicates that I'm saying it is true for any surface attached to $\vec{dl}$
$ \nabla × \vec{E} = -\frac{\partial \vec{B}} {\partial t} $
Taking the divergence of both sides
$0 = \nabla \cdot \frac{\partial \vec{B}} {\partial t} $
$\nabla \cdot \vec{B} = c$
Meaning the condition upon which this is true for any surface is when the divergence of $\vec{B}$ is a constant, which is inline with the condition that $\nabla \cdot \vec{B} = 0$
The intuitive way of showing this, is that  for the surface integral to be independant of the surface, the fluxes of 2 chosen surfaces of the same boundary must be the same, meaning constructing a closed surface out of those 2 surfaces, would change $\vec{da}$ for one of them, meaning the total flux out of the closed surface will be $\phi +(-\phi)$ = 0, using divergence theorem we see that this must be 0.
