Integral over a surface in 4-dimensions Consider the integral of a function $f(x,y,z)$ over a surface embedded in 3 dimensions. 
The surface has a parameterization:
$$g(u,v) = (x(u,v), y(u,v), z(u,v)) $$
The integral is given by:
$$ \iint_{u,v}f(g(u,v))\|\vec{g}_u \times \vec{g}_v\|\,dudv$$
In words, you are placing a curvilinear coordinate system over the surface (a "form-fitting" coordinate system to the surface). You use these curvilinear grid lines to tile the surface with parallelogram area elements. These parallelograms are actually linear approximations to the curvilinear area elements.
Question
Consider a surface integral of a function $f(x,y,z,w)$ over a surface embedded in 4 dimensions. The surface has a parameterization:
$$ g(u,v) = (x(u,v), y(u,v), z(u,v), w(u,v)) $$
The integral is given by
$$ \iint_{u,v} f(g(u,v))(\;\;\;\;)dudv$$
I can embedded parallelograms in 4 dimensions. However my question is, what is the notation used for the linear approximation? The cross product doesn't work in 4-dimensions. So I was wondering if I should go back to something more geometric: $A = bh$. The two side lengths of an infinitesimal parallelogram are given by $\|\vec{g}_u\|du$ and $\|\vec{g}_v\|dv$. The angle between them is given by
$$ \theta = \arccos\Big(\frac{\vec{g}_udu \cdot \vec{g}_vdv}{\|\vec{g}_u\|du\|\vec{g}_v\|dv}\Big) \\
\theta = \arccos\Big(\frac{\vec{g}_u \cdot \vec{g}_v}{\|\vec{g}_u\|\|\vec{g}_v\|}\Big)
$$
And therefore my integral becomes
$$ \iint_{u,v} f(g(u,v)) \|\vec{g}_u\|\|\vec{g}_v\|\sin(\arccos\Big(\frac{\vec{g}_u \cdot \vec{g}_v}{\|\vec{g}_u\|\|\vec{g}_v\|}\Big)) \,dudv$$
Is this even correct? If so, is there simplifying notation for the linear approximation?
 A: Here is another way to think about the quantity $\left| g_u \times g_v \right|$ which will generalize to more than three dimensions. If $g(u,v) = (x(u,v), \, y(u,v), \, z(u,v))$, then the cross product, in components, is
$$ g_u \times g_v = (y_u z_v - y_v z_u, \, z_u x_v - z_v x_u, \, x_u y_v - x_v y_u) $$
Its squared norm is
$$
\begin {align*}
\left| g_u \times g_v \right|^2 &= (y_u z_v - y_v z_u)^2 + (z_u x_v - z_v x_u)^2 + (x_u y_v - x_v y_u)^2 \\
&= (y_u z_v)^2 + (y_v z_u)^2 + (z_u x_v)^2 + (z_v x_u)^2 + (x_u y_v)^2 + (x_v y_u)^2 \\
&\phantom{=} - 2 \left( y_uy_vz_uz_v + z_uz_vx_ux_v + x_ux_vy_uy_v \right)
\end {align*}
$$
A quick computation will convince you that this is equal to 
$$ \left| g_u \right|^2 \left| g_v \right|^2 - \left( g_u \cdot g_v \right)^2 $$
So in conclusion, we can think of this area element quantity as
$$ \left| g_u \times g_v \right| = \sqrt{\left| g_u \right|^2 \left| g_v \right|^2 - \left( g_u \cdot g_v \right)^2} $$
This is also the square root of the determinant of the matrix
$$ G = \left( \begin{array}{cc} g_u \cdot g_u & g_u \cdot g_v \\ g_v \cdot g_u & g_v \cdot g_v \end{array} \right) $$
Now this approach should give you the correct area element for surfaces embedded in higher dimensional space. In your example of a surface in $\Bbb{R}^4$, with $g(u,v) = (x,y,z,w)$, it still makes sense to consider
$$ g_u = (x_u,y_u,z_u,w_u) $$
$$ g_v = (x_v,y_v,z_v,w_v) $$
They are now just 4-dimensional vectors. The area element will still be the square root of the determinant of the matrix $G$ above.
To explain why this approach works you need to know some differential geometry, including the theory of differential forms and Riemannian metrics, so I will just leave it at that.
