Let $U$ be an bounded, open set in $\mathbb{R}^n$ and $\partial U$ be smooth, $u\in C^1 (\bar U)$ and $\nu=(\nu^1,\ldots,\nu^n)$ the outward pointing unit normal vector field. I'm trying to make sense of the integral $$\int_{\partial U}u\nu^idS$$for some fixed $i\in \{1,\ldots,n\}$.
Can some please tell me show me a concrete example (doesn't have to be complicated), so that I can be learning by doing ?
It would be nice if along the way it would also be explained
How is the product "$u\nu^i$" defined ?
What does integrating on $\partial U$ relative to $dS$ mean ? The explanation should be without the use of differential forms, just purely analytically, as I
As soon as I can I'm willing to offer lots of bounty for this, as soon as I can.
Let $U$ be the open unit disk in $\mathbf{R}^{2}$, so that the outward unit normal is $\nu(x, y) = (x, y)$, and $\partial U = S^{1}$ is the unit circle. Parametrize the boundary by $c(t) = (\cos t, \sin t)$ for $0 \leq t \leq 2\pi$. If $u$ is continuously-differentiable on the closed unit disk, your integrals are $$ \int_{S^{1}} ux\, dS = \int_{0}^{2\pi} u(\cos t, \sin t) \cos t\, dt,\qquad \int_{S^{1}} uy\, dS = \int_{0}^{2\pi} u(\cos t, \sin t) \sin t\, dt. $$
One way to interpret $\int_{\partial U} u\nu^{i}\, dS$ is to introduce the vector fields $F = ue_{1} = (u, 0)$ and $G = ue_{2} = (0, u)$. Since $F \cdot \nu = u\nu^{1}$ and $G \cdot \nu = u\nu^{2}$, the integrals are, respectively, the net fluxes of $F$ and $G$ across the unit circle.
Generally, in your set-up you might consider the $n$ vector fields $F_{i} = ue_{i}$, each everywhere parallel to the $i$th coordinate axis. The function product $u\nu^{i} = F_{i} \cdot \nu$ may be interpreted as the normal component of the fluid velocity at a point of $\partial U$, so the integrand $u\nu^{i}\, dS$ represents the flux through a small piece of the boundary, and the integral represents the net flux of $F_{i}$ across the boundary.
To compute such an integral, you'd typically parametrize the boundary of $U$, i.e., find a region $D$ in $\mathbf{R}^{n-1}$ and a continuously-differentiable function $\phi:D \to \partial U$ that is one-to-one (except possibly at boundary points of $D$) and onto.
To form the integrand, compute the $(n - 1)$ partial derivatives $\partial\phi/\partial x_{i}$ of $\phi$ (each partial derivative is a vector-valued function with values in $\mathbf{R}^{n}$), and form their "(generalized) cross product" $$ N = \frac{\partial\phi}{\partial x_{1}} \times \dots \times \frac{\partial\phi}{\partial x_{n-1}}. $$ (Assemble the partials into rows of a matrix, put the standard basis vectors in the last row, and take the determinant, see the Wikipedia formula.) The cross product is a vector-valued function on $D$ (with values in $\mathbf{R}^{n}$). The desired integral is $$ \int_{D} u\bigl(\phi(x)\bigr) N_{i}\, dx_{1} \dots dx_{n-1}, $$ an ordinary integral over a region in $\mathbf{R}^{n-1}$.