Suppose we have a surface S(x,y) = x^2-2y for 0<x<2,0<y<5 , and a magnetic field (2t,3t+1,t).
To find the total magnetic flux through the surface S we find a normal vector to S n = (-Sx,-Sy,1) = n = (-2x,-2,1)
V = -dΦ/dt = 40V
But voltage related to what reference?
As Hosein mentioned, the issue is buried in the difference between electromotive force and voltage.
The scalar potential, which could reasonably be called the voltage field, encodes the curl free part of the electric field -
$$ \vec{E} = -\vec{\nabla} \Phi - \frac{\partial \vec{A}}{\partial t}$$ where $A$ is the magnetic vector potential.
In static problems, $\frac{\partial A}{\partial t}=0$, so the electric field is curl free and the work needed to move a charge $Q$ from $x_i$ to $x_f$ is
$$ Q\int_{\vec{x}_i}^{\vec{x}_f} q \vec{E}\cdot d\vec{l} $$
The answer is the same for any path you take, and is equivalent to $\Phi(x_i) - \Phi(x_f)$.
The same is absolutely not true when dealing with changing magnetic fields, which in general imply nonzero $\frac{\partial A}{\partial t}$. Nonetheless, it's still possible to compute the work per unit charge required to move an object in a closed loop $\gamma$:
$$\frac{W}{Q} = \oint_\gamma d\vec{x} \cdot (\vec{E} + \frac{d\vec{x}}{dt}\times B ) $$
$$ = - \frac{\partial}{\partial t}\oint_\gamma \vec{A}\cdot d\vec{x} = -\frac{\partial}{\partial t} \iint_S B\cdot dA $$
where $S$ is any surface with boundary $\partial S = \gamma$.
We assumed infinitesimally slow (i.e. adiabatic) transport of the charge to throw away the $v\times B$ term.
This quantity has units of volts, but is not a "voltage" in the traditional sense - there's no way to define it independent of the path you took to get from start to finish. It's useful to refer to the voltage as the path-independent part of the work to move a charge from A to B, and the electromotive force to account for the "extra" work you need to do against the induced electric field, which depends strongly on the path the charge takes.
