Vector Integral Calculus
A Brief Introduction to Manifolds and Boundaries
First, some preliminary definitions. I'll talk some about manifolds and boundaries, because, while we're not going to do a full generalized Stokes Theorem here (not yet, at least), I think it is useful to think about the connections between FTC, Fundamental Theorem of Line Integrals, Green's Theorem, Stoke's Theorem, and the Divergence Theorem.
A homeomorphism is a @bijective and continuous function between @topological-spaces that has a continuous @inverse-function.
A manifold is a @topological-space that resembles @euclidean-space near each point. That is, an $n$-dimensional manifold is a topological space with the property that each point has a neighborhood that is homeomorphic to an open subset of $n$-dimensional Euclidean space.
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Another way to put this is that we can approximate a $k$-manifold as closely as we'd like with a $k$-plane. We call this $k$-plane (a line in 1-dimenions, a plane in 2, a half-space in 3, etc) the tangent space.
If $M$ is locally given by a smooth parametrization
$$ \varphi : U \subset \mathbb{R}^k \to M \subset \mathbb{R}^n, $$
and $p = \varphi(u_0)$, then
$$ T_p M = \operatorname{span}\left\{ \frac{\partial \varphi}{\partial u_1}(u_0), \frac{\partial \varphi}{\partial u_2}(u_0), \dots, \frac{\partial \varphi}{\partial u_k}(u_0) \right\}. $$
So it’s the collection of all possible velocity vectors of curves on the manifold passing through $p,$ and is a $k$-dimensional linear subspace of $\mathbb{R}^n.$
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One dimensional manifolds include lines and circles, but not curves that cross themselves. Two dimensional manifolds are also called surfaces, and include planes, discs, torus and more. A solid ball is a 3d manifold.
We will be interested primarily in manifolds with boundaries.
The points on a manifold whose neighborhoods are homeomorphic to a neighborhood in a half $k$-ball form the boundary of the manifold. Formally,
$$ \partial M = \{p \in M | \text{there exists a chart } (U, p) \text{ with } \varphi(p) \in \mathbb{R}^{k-1} \times {0} \subset \mathbb{H}^k \}. $$
That is, $p$ is a boundary point, if, in local coordinates, it maps to the edge of the half-space model.
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The @chart $\varphi$ maps points on the manifold to local coordinates. For example, points on the sphere
$$ S^2 = {(x,y,z) : x^2 + y^2 + z^2 = 1} $$
can be assigned coordinates via a chart (with $N$ representing the north pole) $(U = S^2 \setminus {N}, \varphi)$ with
$$ \varphi(x,y,z) = \left ( \frac{x}{1 - z}, \frac{y}{1 - z} \right ). $$
These local coordinates can then be used to refer to any point on the sphere. So, the definition of a boundary of a manifold above talks about the points that can be mapped to some $k$-vector that has a $0$ in its last dimension. Therefore, the boundary itself is of a dimension $k - 1.$
So, the boundary of a ball $B$, a 3-manifold, is a sphere ($\partial B^3 = S^2$), a 2-manifold, and the boundary of a disc, a 2-manifold, is a circle, a 1-manifold ($\partial D^2 = C^1$).
A simple curve in space is a 1-manifold, and if it is not closed, then its boundary is its endpoints, and they form a 0-manifold.
An closed interval in $R^1$ is a 1-manifold and its boundary (its endpoints) form a 0-manifold.
Not all manifolds have boundaries. A sphere has no boundaries - it is a 2-manifold, and there is nowhere to go with its coordinates that is not part of the sphere. Contrast that with a 3-ball - if we keep moving outward from the center, we'll reach the edge and beyond. A closed curve in space is a 1-manifold without boundary. An open interval in $R^1$ is a manifold without boundary as well, because all points are interior points.
So, given a manifold $M$, $\partial(\partial(M)) = \emptyset.$ Taking the boundary of a $k$-manifold with boundary gives a new $(k-1)$-manifold without boundary. When we view it this way, it becomes apparent that the definition of @boundary-points we would use from topology - points that are limit points of both $M$ and $M^c$ - works for manifolds as well, but we have to be careful to consider the ambient space the manifold exists in. For a manifold to have a boundary, it must exist in an ambient space that contains points not in the manifold, but when we take the manifold's boundary, we reduce the ambient space to only those points in the boundary, and so it is impossible for any boundary points of the boundary to exist.
Line Integrals
We can express line integrals over vector fields generally as follows.
A line integral of a vector function $\vec{F}(\vec{r})$ over a curve $C: \vec{r}(t)$ is defined by
$$ \int_{C} \vec{F}(\vec{r}) \cdot d \vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt \tag{a} $$
where $\vec{r}(t)$ is the parametric representation of $C.$
Writing (a) in terms of components, with $d \vec{r} = [dx, dy, dz]$ and $' = d/dt,$ we get
$$ \int_{C} \int_{C} \vec{F}(\vec{r}) \cdot d \vec{r} = \int_{C} (F_1 dx + F_2 dy + F_3 dz) = \int_{a}^{b} (F_1 x' + F_2 y' + F_3 z') dt). $$
Surface Integrals
Given a piecewise smooth surface $S,$ we can parameterize it as
$$ \vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle = x(u,v) \vec{i} + y(u,v) \vec{j} + z(u,v) \vec{k}, $$
with $u_0 \leq u < u_1$ and $v_0 \leq v \leq v_1$ (i.e. $u$ and $v$ vary over a region $R$ in the $uv$-plane).
Now, $S$ has a @normal-vector and unit @normal-vector, respectively, as
$$ \vec{N} = \vec{r}_u \times \vec{r}_v, \quad \vec{n} = \frac{1}{|\vec{N}|} \vec{N} $$
at every point, except perhaps for some edges or cusps, such as for cubes and cones. For a given vector function $\vec{F}$ we can now define the surface integral over $S$ by
$$ \iint_S \vec{F} \cdot \vec{n} dA = \iint_R \vec{F}(\vec{r}(u,v)) \cdot \vec{N}(u,v) du dv. $$
Volume Integrals
If $T$ is a closed, bounded, three-dimensional region of space, and $f(x,y,z)$ is defined and continuous in a domain containing $T,$ then the volume integral of $f(x,y,z)$ over the region $T$ is denoted by
$$ \iiint_T f(x,y,z) dx dy dz = \iiint_T f(x,y,z) dV. $$
Such integrals can be evaluated via three successive integrations.
Note: for cylindrical coordinates, we need to include $r$ as a scaling factor in the integrand:
$$ \iiint_T f\,dV =\int_{z_0}^{z_1}\!\int_{\theta_0}^{\theta_1}\!\int_{r_0(\theta,z)}^{\,r_1(\theta,z)} f(r\cos\theta,r\sin\theta,z)\, r\, dr\, d\theta\, dz. $$
Divergence Theorem of Gauss
The Divergence Theorem allows us to translate between a volume integral over an enclosed region and a surface integral over the surface enclosing that region, and vice-versa.
Let $T$ be a closed bounded region in a space whose boundary is a @piecewise smooth @orientable surface $S.$ Let $\vec{F}(x,y,z)$ be a vector function that is continuous and has continuous first partial derivatives in some domain containing $T.$ Then
$$ \iiint_T \div{\vec{F}} dV = \iint_S \vec{F} \cdot \vec{n} dA. $$
Stoke's Theorem
Stoke's Theorem allows us to translate between a line integral around a boundary of a surface and an integral over that surface.
Let $S$ be a @piecewise smooth oriented surface in space and let the boundary of $S$ be a piecewise smooth simple closed curve $C.$ Let $\vec{F}(x,y,z)$ be a continuous vector function that has continuous first partial derivatives in a domain in space containing $S.$ Then
$$ \iint_S (\curl{\vec{F}}) \cdot \vec{n} dA = \oint_C \vec{F} \cdot \vec{r'}(s) ds. \tag{2} $$
Here $\vec{n}$ is a unit @normal-vector of $S.$ $\vec{r'} = d \vec{r}/ds$ is the unit tangent vector and $s$ the arc length of $C.$
In components, (2) becomes
$$ \iint_R \left [ \left ( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}\right ) N_1 + \left ( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}\right ) N_2 + \left ( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right ) N_3 \right ] du dv \\ = \oint_{\overline{C}} (F_1 dx + F_2 dy + F_3 dz). $$
Here, $ \vec{F} = [F_1, F_2, F_3], \vec{N} = [N_1, N_2, N_3], \vec{n}dA = \vec{N} du dv, \vec{r'} ds = [dx, dy, dz],$ and $R$ is the region with boundary curve $\overline{C}$ in the $uv$-plane corresponding to $S$ represented by $\vec{r}(u,v).$