Example of differential form usage of Stoke's theorem There are many examples that show how Kelvin-Stokes theorem is used. 
But I would like to see an example that uses differential form usage of Stoke's theorem and is hard or impossible to solve by Kelvin-Stokes theorem.
Can anyone present such example and show how they are solved?
 A: I am not sure what you are expecting; offhand, one can cook up examples of applying Stokes' Theorem in the language of Differential Forms, though I am not sure how they would be "impossible" in the Classical Formulation - the generalized one in Forms 
$$
\int_{\mathfrak{S}} {\mathrm{d}\omega} = \int_{\partial \mathfrak{S}} {\omega}
$$
over an orientable manifold $\mathfrak{S}$ states exactly the same thing as the Classical Kelvin-Stokes' Theorem:
$$
\iiint_{\mathbf{S}} {\left( \nabla \times \mathbf{F} \right) \cdot \mathbf{n} \, \mathrm{d} \mathbf{S}} = \int_{\partial \mathbf{S}} {\mathbf{F} \cdot \mathrm{d} \mathbf{r}} 
$$
when applied to vector fields over oriented "surfaces".
However, I can display an example of a computation of the integral over some manifold employing the General Stokes' Theorem.
Consider the following: Let $\omega$ be the form $x \mathrm{d} y \wedge \mathrm{d} z + y \mathrm{d} z \wedge \mathrm{d} x + z \mathrm{d} x \wedge \mathrm{d} y$, and 
$$
\mathbf{M} = \left\{ \left( x,y,z \right) \, \Big| \, x^2 + y^2 \leq 4, \, 0 \leq z \leq 4 \right\},
$$
the solid cylinder of radius 2 extruding along the $z$ axis in $\mathbb{R}^3$. It is immediate to see that $\partial \mathbf{M}$ is the collection of points $(x,y,z)$ satisfying 
$$
x^2 + y^2 = 4, \qquad 0 \leq z \leq 4.
$$
Indeed, as we are dealing with a Cylinder, it will be fruitful to consider our Manifold $\mathbf{M}$ in Cylindrical Coordinates. Therefore, we have that 
\begin{align}
\mathbf{M} &= \left\{ \left( x,y,z \right) \, \Big| \, x^2 + y^2 \leq 4, \, 0 \leq z \leq 4 \right\} \\
&= \left\{ \left( r,\theta,z \right) \, \Big| \, 0 \leq r \leq 2, \, 0 \leq \theta \leq 2 \pi, \, 0 \leq z \leq 4 \right\} .
\end{align}
Supposing that we want to know the value of the form $\omega$ over the boundary of $\mathbf{M}$, it will be much more direct to calculate the corresponding value using Stoke's Theorem.
Now, we have by Stokes' Theorem that 
\begin{align}
\int_{\partial \mathbf{M}} {\omega} &= \int_{\mathbf{M}} {\mathrm{d}\omega} \\
&= \iiint \limits_{\mathbf{M}} {3 \mathrm{d} x \wedge \mathrm{d} y \wedge \mathrm{d} z \left( \gamma_r \gamma_\theta \gamma_z \right) \, \left| \mathrm{d} r \mathrm{d} \theta \mathrm{d} z \right| }
\end{align}
where $D \gamma = \overbrace{ \left[ \begin{array}{ccc}
\cos \theta & -r \sin \theta & 0 \\
\sin \theta & r \cos \theta & 0 \\
0 & 0 & 1
\end{array} \right] }^{\gamma_r \qquad \quad \gamma_\theta \qquad \gamma_z}$ is the matrix formed of tangents operating on the map 
$$\gamma(x,y,z) = \left( x(r, \theta, z), y(r, \theta, z), z(r, \theta, z) \right) = \left( r \sin \theta, r \cos \theta, z \right)
$$
that performs the change of variables into Cylindrical Coordinates. Thus, by the definition of a 3-form acting upon the collection of "vectors" $\gamma_r \gamma_\theta \gamma_z$, we have that 
\begin{align}
\int_{\partial \mathbf{M}} {\omega} &= 3 \iiint \limits_{\mathbf{M}} { \left| \begin{array}{ccc}
\cos \theta & -r \sin \theta & 0 \\
\sin \theta & r \cos \theta & 0 \\
0 & 0 & 1
\end{array} \right| 
\mathrm{d} r \mathrm{d} \theta \mathrm{d} z } \\
%
%
% second line
%
%
&= 3 \iiint \limits_{\mathbf{M}} { \left| \begin{array}{cc}
\cos \theta & -r \sin \theta \\
\sin \theta & r \cos \theta \\
\end{array} \right| 
\mathrm{d} r \mathrm{d} \theta \mathrm{d} z } \\
%
%
% third line
%
%
&= 3 \left( \int_0 ^2 r \mathrm{d} r \right) \left( \int_0 ^{2 \pi} \mathrm{d} \theta \right) \left( \int_0 ^4 \mathrm{d} z \right) = 48 \pi .
\end{align}
Similarly, one can verify the theorem by computing the integral $\int_{\partial \mathbf{M}} {\omega}$ directly.
Hopefully this was helpful, and at least somewhat along the lines of what you were looking for.
