[proofplan] By the definition of the integral of a $2$-form over an oriented parametrised surface (see [Integration of Differential Forms](/theorems/1529)), $\int_S \omega_F = \int_D r^* \omega_F$. We compute the pullback $r^* \omega_F$ explicitly: pulling back the coordinate $1$-forms $dx, dy, dz$ yields linear combinations of $du$ and $dv$ whose wedge products produce $2 \times 2$ Jacobian minors. Collecting the three terms in $\omega_F$, the coefficient of $du \wedge dv$ is precisely the $3 \times 3$ determinant whose rows are $F(r)$, $r_u$, and $r_v$ — that is, the scalar triple product $F(r) \cdot (r_u \times r_v)$. The result follows by interpreting $\int_D h\, du \wedge dv$ as the [Lebesgue integral](/page/Lebesgue%20Integral) $\int_D h\, d\mathcal{L}^2$, which is valid because the parametrisation is orientation-compatible. [/proofplan]

[step:Reduce the surface integral to a pullback integral over the parameter domain]By the definition of the integral of a differential form over an oriented parametrised submanifold (see [Integration of Differential Forms](/theorems/1529)), for any smooth $2$-form $\omega \in \Omega^2(U)$ on an [open set](/page/Open%20Set) $U \supseteq S$ and any orientation-compatible smooth parametrisation $r: D \to S$, \begin{align*} \int_S \omega = \int_D r^* \omega. \end{align*} The hypotheses are met: $r \in C^\infty(D; \mathbb{R}^3)$ is an immersion by assumption (so it is a smooth parametrisation), $S = r(D) \subseteq U$ lies in the domain of $\omega_F$, and the parametrisation is orientation-compatible by hypothesis. Applying this with $\omega = \omega_F$ gives \begin{align*} \int_S \omega_F = \int_D r^* \omega_F. \end{align*} It therefore suffices to compute $r^* \omega_F$ as a $2$-form on $D$.[/step]

[guided]The integral of a $2$-form over an oriented surface is *defined* via parametrisation: we pull the form back to a region of $\mathbb{R}^2$ and integrate the resulting top-degree form there. The well-definedness of this construction — independence of the choice of orientation-compatible parametrisation — is itself part of [Integration of Differential Forms](/theorems/1529); we do not need to reprove it here. To invoke that theorem we must verify three things: - $r$ is a smooth parametrisation. Indeed $r \in C^\infty$ and is an immersion by hypothesis. - The image lies in the domain of the form. We have $S = r(D) \subseteq U$ by hypothesis, and $\omega_F$ is defined on all of $U$. - The parametrisation respects orientation. This is precisely the orientation-compatibility hypothesis: $r$ pushes $du \wedge dv$ forward to the chosen orientation of $S$. With these conditions verified, the theorem yields \begin{align*} \int_S \omega_F = \int_D r^* \omega_F, \end{align*} and the entire problem reduces to computing the pullback $r^* \omega_F$ as a $2$-form on $D$.[/guided]

[step:Pull back the coordinate $1$-forms $dx$, $dy$, $dz$ along $r$] Write $r = (x, y, z)$ where $x, y, z \in C^\infty(D)$. The pullback of a coordinate $1$-form along a smooth map equals the differential of the corresponding component function. Thus \begin{align*} r^* dx &= dx \circ r = x_u\, du + x_v\, dv, \\ r^* dy &= y_u\, du + y_v\, dv, \\ r^* dz &= z_u\, du + z_v\, dv, \end{align*} where subscripts denote partial derivatives, e.g. $x_u := \partial_u x$. [/step]

[step:Compute the pullbacks of the basis $2$-forms $dy \wedge dz$, $dz \wedge dx$, $dx \wedge dy$]Pullback commutes with the wedge product: for any smooth map $r$ and forms $\alpha, \beta$, one has $r^*(\alpha \wedge \beta) = (r^*\alpha) \wedge (r^*\beta)$. We compute each of the three basis $2$-forms in turn, using the bilinearity of $\wedge$, the antisymmetry $du \wedge dv = -dv \wedge du$, and the vanishing $du \wedge du = dv \wedge dv = 0$. For the first: \begin{align*} r^*(dy \wedge dz) &= (y_u\, du + y_v\, dv) \wedge (z_u\, du + z_v\, dv) \\ &= y_u z_v\, du \wedge dv + y_v z_u\, dv \wedge du \\ &= (y_u z_v - y_v z_u)\, du \wedge dv. \end{align*} By the identical computation with $(y, z)$ replaced by $(z, x)$ and then by $(x, y)$: \begin{align*} r^*(dz \wedge dx) &= (z_u x_v - z_v x_u)\, du \wedge dv, \\ r^*(dx \wedge dy) &= (x_u y_v - x_v y_u)\, du \wedge dv. \end{align*} These three coefficients are exactly the components of the cross product $r_u \times r_v$. Recalling \begin{align*} r_u = (x_u, y_u, z_u), \qquad r_v = (x_v, y_v, z_v), \end{align*} the standard determinant formula for the cross product gives \begin{align*} r_u \times r_v = \bigl( y_u z_v - y_v z_u,\ z_u x_v - z_v x_u,\ x_u y_v - x_v y_u \bigr). \end{align*} Denoting the three components by $(r_u \times r_v)_1, (r_u \times r_v)_2, (r_u \times r_v)_3$, we have shown \begin{align*} r^*(dy \wedge dz) &= (r_u \times r_v)_1\, du \wedge dv, \\ r^*(dz \wedge dx) &= (r_u \times r_v)_2\, du \wedge dv, \\ r^*(dx \wedge dy) &= (r_u \times r_v)_3\, du \wedge dv. \end{align*}[/step]

[guided]Each basis $2$-form on $\mathbb{R}^3$ pulls back to a scalar multiple of $du \wedge dv$, because $D$ is two-dimensional and $du \wedge dv$ spans $\Lambda^2 T_{(u,v)}^* D$. The scalar that appears is, in each case, a $2 \times 2$ Jacobian minor of $r$. The pattern is clearest if we view it as a determinant. For two smooth functions $f, g: D \to \mathbb{R}$, \begin{align*} df \wedge dg = (f_u\, du + f_v\, dv) \wedge (g_u\, du + g_v\, dv) = \det \begin{pmatrix} f_u & g_u \\ f_v & g_v \end{pmatrix} du \wedge dv = (f_u g_v - f_v g_u)\, du \wedge dv. \end{align*} This is the *Jacobian determinant of the pair $(f, g)$*, sometimes written $\frac{\partial(f,g)}{\partial(u,v)}$. Applying this with $(f, g) = (y, z)$, $(z, x)$, $(x, y)$ produces the three minors \begin{align*} \frac{\partial(y, z)}{\partial(u, v)}, \qquad \frac{\partial(z, x)}{\partial(u, v)}, \qquad \frac{\partial(x, y)}{\partial(u, v)}, \end{align*} which are precisely the entries of the cross product \begin{align*} r_u \times r_v = \det \begin{pmatrix} e_1 & e_2 & e_3 \\ x_u & y_u & z_u \\ x_v & y_v & z_v \end{pmatrix} \end{align*} expanded along the first row. (Note the cyclic order $(y,z), (z,x), (x,y)$ — not $(x,y), (x,z), (y,z)$ — which is precisely what makes $\omega_F$ encode the cross product rather than some other combination.)[/guided]

[step:Assemble the pullback $r^* \omega_F$ as a scalar triple product]By the linearity of pullback, \begin{align*} r^* \omega_F &= (F_1 \circ r)\, r^*(dy \wedge dz) + (F_2 \circ r)\, r^*(dz \wedge dx) + (F_3 \circ r)\, r^*(dx \wedge dy). \end{align*} Substituting the three pullbacks computed in the previous step, \begin{align*} r^* \omega_F &= \Bigl[ (F_1 \circ r)\,(r_u \times r_v)_1 + (F_2 \circ r)\,(r_u \times r_v)_2 + (F_3 \circ r)\,(r_u \times r_v)_3 \Bigr]\, du \wedge dv \\ &= \bigl( F(r(u,v)) \cdot (r_u \times r_v) \bigr)\, du \wedge dv. \end{align*} Here the bracket is the Euclidean dot product in $\mathbb{R}^3$ of the vectors $F(r(u,v))$ and $(r_u \times r_v)(u,v)$, which is the scalar triple product of $F(r)$, $r_u$, $r_v$.[/step]

[guided]We have now identified the pullback $r^* \omega_F$ as a single scalar function on $D$ times the standard area form $du \wedge dv$. The scalar function is the scalar triple product \begin{align*} F(r) \cdot (r_u \times r_v) = \det \begin{pmatrix} F_1(r) & F_2(r) & F_3(r) \\ x_u & y_u & z_u \\ x_v & y_v & z_v \end{pmatrix}, \end{align*} which is the geometric quantity computing the signed volume of the parallelepiped spanned by $F(r)$, $r_u$, $r_v$. Equivalently, since $r_u \times r_v$ is normal to $S$ with magnitude equal to the local area scaling factor, $F \cdot (r_u \times r_v)$ measures the normal component of $F$ weighted by infinitesimal surface area — exactly the integrand we expect for a flux.[/guided]

[step:Identify the parameter-domain integral as a Lebesgue integral over $D$]By orientation-compatibility of $r$, the form $du \wedge dv$ represents the positive orientation on $D$ induced from the orientation of $S$. By definition of the integral of a top-degree form on an oriented open subset of $\mathbb{R}^2$ (again via [Integration of Differential Forms](/theorems/1529)), for any continuous scalar $h: D \to \mathbb{R}$ for which the integral exists, \begin{align*} \int_D h\, du \wedge dv = \int_D h(u, v)\, d\mathcal{L}^2(u, v). \end{align*} Combining the results of the previous steps, \begin{align*} \int_S \omega_F = \int_D r^* \omega_F = \int_D \bigl( F(r(u, v)) \cdot (r_u(u, v) \times r_v(u, v)) \bigr)\, d\mathcal{L}^2(u, v), \end{align*} which is the asserted identity.[/step]

[guided]The last equality replaces the "$\,du \wedge dv$" notation, which lives in the theory of differential forms, with "$\,d\mathcal{L}^2(u, v)$", which lives in the theory of Lebesgue integration. The reason this substitution is legitimate is precisely the orientation-compatibility hypothesis: the integral of a top-degree form against the standard orientation of $\mathbb{R}^2$ is *defined* to be the [Lebesgue integral](/page/Lebesgue%20Integral) of its coefficient function. If the parametrisation reversed the orientation, an extra sign would appear; the hypothesis rules this case out. Chaining the three equalities: - $\int_S \omega_F = \int_D r^* \omega_F$ (definition of the integral of a $2$-form on a parametrised surface, Step 1); - $r^* \omega_F = \bigl( F(r) \cdot (r_u \times r_v) \bigr)\, du \wedge dv$ (pullback computation, Steps 2–4); - $\int_D h\, du \wedge dv = \int_D h\, d\mathcal{L}^2$ (top-degree forms on $\mathbb{R}^2$ with the standard orientation), we obtain the desired identity. This completes the proof.[/guided]