Let $\Gamma$ be a closed smooth surface in $\mathbb{R}^{3}$, and $\mu:\Gamma\rightarrow\mathbb{R}$. We assume $\Gamma$ can be parametrized in $(u,v)$ such that $\mathbf{x}(u,v)\in\Gamma$ for $(u,v)\in [u_a,u_b]\times[v_a,v_b]$, where $\mathbf{x}(u,v) = (x(u,v),y(u,v),z(u,v))$. Consider the following surface integral $$\int\limits_{\Gamma}\mu(\mathbf{x})dS_{\mathbf{x}} = \int\limits_{u_a}^{u_b}\int\limits_{v_a}^{v_b}\mu(\mathbf{x}(u,v))\|\mathbf{x}_{u}\times\mathbf{x}_{v}\|dudv.$$
Now, we expand $\mu$ in a Taylor series (for simplicity only to first order) in $u$ and $v$ around $(u_{0},v_{0})$, and get $$\int\limits_{u_a}^{u_b}\int\limits_{v_a}^{v_b}(\mu(\mathbf{x}(u_{0},v_{0})) + \alpha(u_{0},v_{0})(u-u_{0}) + \beta(u_{0},v_{0})(v-v_{0}))\|\mathbf{x}_{u}\times\mathbf{x}_{v}\|dudv.$$ Here, $\alpha$ and $\beta$ are factors in two first terms in the multivariate Taylor expansion of $\mu$. My question is, what are the explicit expressions for $\alpha$ and $\beta$?
My suggestion (which is incorrect) is that $\alpha(u_{0},v_{0}) = \frac{\partial \mu(u_{0},v_{0})}{\partial u}\frac{1}{\|\mathbf{x}_{u}\times\mathbf{x}_{v}\|}$ and $\beta(u_{0},v_{0}) = \frac{\partial \mu(u_{0},v_{0})}{\partial v}\frac{1}{\|\mathbf{x}_{u}\times\mathbf{x}_{v}\|}$. The (incorrect) motivation is as follows. If $\Gamma$ is a curve in $\mathbb{R}^{2}$, parametrized in $t$ such that $\mathbf{x}(t)\in\Gamma$, then the first order term in the Taylor series expansion of $\mu$ around $t_{0}$ is $$\mu^{\prime}(\mathbf{x}(t_{0})) = \frac{d\mu(\mathbf{x}(t_{0}))}{dS_{\mathbf{x}}}(t-t_{0}) = \frac{d\mu(\mathbf{x}(t_{0}))}{dt}\frac{dt}{dS_{\mathbf{x}}}(t-t_{0}) = \frac{d\mu(\mathbf{x}(t_{0}))}{du}\frac{1}{\|x^{\prime}(t_{0})\|},$$ since $dS_{\mathbf{x}}/dt = \|x^{\prime}\|$.
I apologize if my question is unclear. I will happily update and modify to make it more clear.