MTW Gravitation Chap. 4. ELECTROMAGNETISM AND DIFFERENTIAL FORMS Pg 100
In differential forms, the $\bf{Faraday}$ has the definition: $$ \mathbf{F}= \frac{1}{2} F_{\alpha \beta} \mathbf{d}x^{\alpha} \wedge \mathbf{d}x^{\beta} \tag{1}$$ For an infinitesimal surface $\bf{u \wedge v}$ and for general $\bf{F}$ $$\int_{surface} \mathbf{F}=\int_{\mathbf{u} \wedge \mathbf{v}} \mathbf{F}=\mathbf{F(\bf{u,v})} \tag{2}$$ But I do not see how the second equality in (2) holds.
My attempt to derive the result:
Parameterise the surface by generic points $$\mathcal{P}=\mathcal{P}(\lambda^1, \lambda^2) \tag{3}$$
Using the computational rule for integration:
$$\int_{surface} \mathbf{F}= \int <\mathbf{F}, \frac{\partial \mathcal{P}}{\partial \lambda^1} \wedge \frac{\partial \mathcal{P}}{\partial \lambda^2}>d\lambda^1 d\lambda^2 \tag{4}$$
Since $$ \frac{\partial \mathcal{P}}{\partial \lambda^1} \wedge \frac{\partial \mathcal{P}}{\partial \lambda^2}= \mathbf{u \wedge v} \tag{5}$$
(4) becomes $$\int <\mathbf{F}, \mathbf{u} \wedge \mathbf{v}>d\lambda^1 d\lambda^2 \tag{6}$$
Expressing (6) in terms of its components, I got $$\int <\frac{1}{2} F_{\alpha \beta} \mathbf{d}x^{\alpha} \wedge \mathbf{d}x^{\beta}, u^{\rho}\mathbf{e_{\rho}} \wedge v^{\sigma}\mathbf{e_{\sigma}}>d\lambda^1 d\lambda^2 \tag{7}$$
Or equivalently $$\int \frac{1}{2} F_{\alpha \beta} u^{\rho}v^{\sigma}<\mathbf{d}x^{\alpha} \wedge \mathbf{d}x^{\beta}, \mathbf{e_{\rho}} \wedge \mathbf{e_{\sigma}}>d\lambda^1 d\lambda^2 \tag{8}$$ which simplifies to $$\int \frac{1}{2} F_{\alpha \beta} u^{\rho}v^{\sigma} \delta^{\alpha \beta}_{\rho \sigma}d\lambda^1 d\lambda^2= \int \frac{1}{2} F_{\alpha \beta} u^{\alpha}v^{\beta} d\lambda^1 d\lambda^2\tag{9}$$ and is the closest expression to $$\mathbf{F}(\mathbf{u,v})=F_{\alpha \beta}u^{\alpha}v^{\beta} \tag{10}$$ that I can get.
How can I proceed from (9) to obtain (2), or rather how can I remove the integral and the 1/2 factor in (9)? Or is there any misconception in my working?
Two points: First, we are assuming $F_{\alpha\beta} = -F_{\beta\alpha}$; the factor of $1/2$ is there because we are summing over all $\alpha,\beta$ rather than over $\alpha<\beta$. At the end, you are missing terms: You should in fact have $u^\alpha v^\beta - u^\beta v^\alpha$, and this, together with skew symmetry of $F$, will get rid of the $1/2$.
Second, when you parametrize the parallelogram, you have $0\le \lambda^1,\lambda^2 \le 1$, so the double integral of the constant over the unit square gives you what you want. (Remember that they are being intuitive here, and on an "infinitesimal surface" they treat the coefficients $F_{\alpha\beta}$ as constants.)