Consider the two form
\begin{equation*}
\omega(x, y, z) = P(x, y, z) \, dy\wedge dz + Q(x, y, z)\, dz\wedge dx + R(x, y, z) \, dx\wedge dy
\end{equation*}
in \(\bbR^{3}\text{.}\) Suppose \(\vg: (u, v)\in I^{2} \mapsto \bbR^{3}\) is a singular \(2\)-cube. We defined \(\int_{\vg}\omega \) as \(\int_{I^{2}} \vg^{*}(\omega)\text{.}\) Let’s see a more concrete form of \(\vg^{*}(\omega)\text{.}\)
\begin{align*}
\vg^{*}(\omega) \amp= P\circ \vg \left(\frac{\partial y}{\partial u} \, du + \frac{\partial y}{\partial v} \, dv\right)
\wedge \left(\frac{\partial z}{\partial u} \, du + \frac{\partial z}{\partial v} \, dv\right) \\
\amp + Q\circ \vg \left(\frac{\partial z}{\partial u} \, du + \frac{\partial z}{\partial v} \, dv\right)
\wedge \left(\frac{\partial x}{\partial u} \, du + \frac{\partial x}{\partial v} \, dv\right) \\
\amp + R\circ \vg \left(\frac{\partial x}{\partial u} \, du + \frac{\partial x}{\partial v} \, dv\right)
\wedge \left(\frac{\partial y}{\partial u} \, du + \frac{\partial y}{\partial v} \, dv\right) \\
\amp= P\circ \vg \left(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right) du\wedge dv \\
\amp+ Q\circ \vg \left(\frac{\partial z}{\partial u} \frac{\partial x}{\partial v}-
\frac{\partial z}{\partial v} \frac{\partial x}{\partial u} \right) du\wedge dv \\
\amp+ R\circ \vg \left(\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}-
\frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \right) du\wedge dv
\end{align*}
Note that
\begin{equation*}
D_{u}\vg\times D_{v}\vg=\left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u}, \frac{\partial z}{\partial u} \frac{\partial x}{\partial v}-
\frac{\partial z}{\partial v} \frac{\partial x}{\partial u},
\frac{\partial x}{\partial u} \frac{\partial y}{\partial v}-
\frac{\partial x}{\partial v} \frac{\partial y}{\partial u}\right),
\end{equation*}
so
\begin{equation*}
\vg^{*}(\omega)=(P\circ \vg, Q\circ \vg, R\circ \vg) \cdot \left( D_{u}\vg\times D_{v}\vg\right)\,
du\wedge dv,
\end{equation*}
and if we use \(\nu\) to denote the unit vector in the direction of \(D_{u}\vg\times D_{v}\vg\) (when it is not \(\mathbf 0\)), then
\begin{equation*}
\vg^{*}(\omega)=(P\circ \vg, Q\circ \vg, R\circ \vg)\cdot \nu \Vert D_{u}\vg\times D_{v}\vg\Vert
\, du\wedge dv
\end{equation*}
and
\begin{equation*}
\int_{I^{2}} \vg^{*}(\omega)= \int_{I^{2}} (P\circ \vg, Q\circ \vg, R\circ \vg)\cdot \nu \Vert D_{u}\vg\times D_{v}\vg\Vert \, du\, dv.
\end{equation*}
Recall that the integral of \(\Vert D_{u}\vg\times D_{v}\vg\Vert\) gives the surface area of the parametric surface \(\vg=\vg(u, v)\) over the parameter domain, and the integral on the right hand side above is identified as the surface integral
\begin{equation*}
\int_{\vg(I^{2})} (P, Q, R)\cdot \nu \, dA.
\end{equation*}
Because of this relation one often finds in older texts that
\(\int_{S} (P, Q, R)\cdot \nu \, dA\) is written as
\(\int_{S} P\, dy dz + Q\, dz dx +R\, dx dy\) on a surface
\(S\text{.}\)
If we look at the term involving \(P, Q\) or \(R\) individually, say, the term \(P\circ \vg \left(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right)\text{,}\) then
\begin{equation*}
\int_{I^{2}} P\circ \vg \left(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right) \, du dv
=\int_{\text{Projection into the } y, z
\text{ plane of } \vg(I^{2})} P \,dy dz
\end{equation*}
according to the change of variables formula in integration, provided that \(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u}>0\text{.}\) In other words, this integral can be treated as an integral in the \(y\mbox{--}z\) coordinate plane. When the sign is not specified, the integral takes into account of the orientation of the projection, such as in the case that \(\vg\) is a parametrization for the round sphere that includes portions of both the upper and lower hemisphere, where \(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u}\) will be positive in some region while negative in some other region, so the resulting integral reflects this. As a result, \(\int_{\vg}\omega\) can be interpreted as projecting \(\omega\) to each of the possible two dimensional coordinate planes and computing the integral of the projected component in the projected domain, taking into account of the orientation of the projection, and summing up these integrals.
For example, if \(\vg\) is a parametrization for the upper half unit sphere centered at the origin, then its projection onto the \(y\mbox{--}z\) plane will be two-to-one, with the right half and left half carrying opposite signs in the Jacobian. Suppose the unit normal corresponding to the parametrization \(\vg\) is outward with respect to the sphere. Since \(\frac{\partial y}{\partial u} \frac{\partial z}{\partial v}-
\frac{\partial y}{\partial v} \frac{\partial z}{\partial u}\) is the \(x\)-component of the normal, we know it is positive when \(\vg\) lands on the right half of the sphere and is negative and when \(\vg\) lands on the left half of the sphere. If \(\omega=x\, dy\wedge dz\text{,}\) then the \(x\) factor would also take opposite signs on the two halves of the sphere, so
\begin{equation*}
\int_{\vg: x=\vg_{1}\ge 0} x\, dy\wedge dz =\int_{y^{2}+z^{2}\le 1, z\ge 0} \sqrt{1-y^{2}-z^{2}}\, dy dz\text{,}
\end{equation*}
\begin{equation*}
\int_{\vg: x=\vg_{1}\le 0} x\, dy\wedge dz =\int_{y^{2}+z^{2}\le 1, z\ge 0} -\sqrt{1-y^{2}-z^{2}} (-1)\, dy dz\text{,}
\end{equation*}
which results in
\begin{equation*}
\int_{\vg}x\, dy\wedge dz=2 \int_{y^{2}+z^{2}\le 1, z\ge 0} \sqrt{1-y^{2}-z^{2}}\, dy dz,
\end{equation*}
provided the parametrization gives a positive Jacobian when \(x >0\) (which is the case if \(D_{u}\vg\times D_{v}\) points upward).
If one chooses to use \(\vg(x,y)=(x,y, \sqrt{1-x^{2}-y^{2}})\) as a parametrization, then one could also use this parametrization to evaluate the integral directly, using
\begin{equation*}
dy\wedge dz= dy\wedge (\frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy)=-
\frac{\partial z}{\partial x}dx\wedge dy,
\end{equation*}
and get
\begin{equation*}
\int_{\vg}x\, dy\wedge dz= - \int_{x^{2}+y^{2}\le 1} x \frac{\partial z}{\partial x}\,dx\, dy
=\int_{x^{2}+y^{2}\le 1}\frac{x^{2}}{\sqrt{1-x^{2}-y^{2}}} \,dx\, dy.
\end{equation*}
A third option to evaluate the above integral is to think of \(\vg\) as part of \(\partial c\) for some singular \(3\)-cube (or chain). One could imagine defining a \(3\)-cube \(c\) mapping into the upper half unit ball such that its restriction to its top face gives rise to \(\vg\text{,}\) its restriction to its bottom face gives rise to a parametrization for the unit disk in the plane \(z=0\text{,}\) and its restriction to its lateral faces maps to the unit circle \(x^{2}+y^{2}=1, z=0\text{,}\) then
\begin{equation*}
\int_{\partial c} x \, dy\wedge dz= \int_{\vg}x\, dy\wedge dz - \int_{\text{bottom}} x\, dy\wedge dz.
\end{equation*}
But on the bottom face, \(z=0\text{,}\) so \(dy\wedge dz=0\) in the integral. On the other hand, we can apply the Stokes’ Theorem to get
\begin{equation*}
\int_{\vg}x\, dy\wedge dz= \int_{c} d(x \, dy\wedge dz)= \int_{c} dx \wedge dy\wedge dz.
\end{equation*}
A similar consideration for how to compute integrals of a \(3\)-form leads us to conclude that \(\int_{c} dx \wedge dy\wedge dz\) is the volume of the upper half unit sphere.
In higher dimensions, a surface may no longer have a notion of a normal vector, but the latter interpretation of projecting on each coordinate plane still applies.
