Integral over a hypersurface In Landau-Lifschitz(Volume 2) is given
Integral over a hypersurface, i.e. over a three-dimensional manifold. In three dimensional space the volume of the parallelepiped spanned by three vectors is equal to the determinant of the third rank formed from the components of the vectors. One obtains analogously the projections of the volume of the parallelepiped (i.e. the "areas" of the hypersurface) spanned by three four-vectors $du^i, dv^i, dw^i;$ they are given by determinants 
\begin{equation*}
dS^{ikl}=
\begin{vmatrix}
du^i & dv^i & dw^i\\
du^k & dv^k & dw^k\\
du^l & dv^l & dw^l
\end{vmatrix}
\end{equation*}
which form a tensor of rank 3, antisymmetric in all three indices.
If it is three-dimensional manifold why it is called a hypersurface? What is actually a hypersurface? How are the projections of the volume of the parallelepiped can be the "areas" of the hypersurface?
 A: A hypersurface is another name for a codimension-1 manifold. That is, it is a manifold whose dimension is one less than that of the ambient space. In 3D this means a hypersurface is a (3-1)=2D manifold ie a regular 'surface'.
Hypersurfaces arise frequently as the boundaries of manifolds. The boundary of a manifold-with-boundary is a manifold (without boundary) of one lower dimension than the original manifold. For example the boundary of the 3-ball $B^3=\{(x,y,z)|x^2+y^2+z^2\leq 1\}$ is the 2-sphere $S^2=\{(x,y,z)|x^2+y^2+z^2=1\}$.
The words 'volume' and 'area' are themselves confusing as they can mean either 3-dimensional (or 2-dim respectively) measures OR they can be used in the sense that volume is the 'top' measure. In this sense, an integral over a 4D region of 4D spacetime is a 'volume integral'. Since 4D regions are bounded by 3D hypersurfaces, the 'surface area' is given by some triple integral, but it still makes sense to call it an area, not a volume.
One setting where this all comes up is the general case of Stoke's theorem, which states that a volume integral of the derivative of a function can be related to a surface integral of that function. This theorem holds in any number of dimensions and it would be silly to have to give them all names, so we tend to call the highest dimensional integral a 'volume' and the next highest a 'surface area'.
