Stokes' theorem is a fundamental result in vector calculus that relates the flux of a vector field across a surface to the circulation of the same vector field along the boundary curve of that surface. The theorem is named after Sir George Gabriel Stokes, who formulated it.

Stokes' theorem can be stated in general terms as follows:

Let $\mathbf{F}$ be a vector field defined on an open region in three-dimensional space, and let $S$ be a smooth, oriented surface that is bounded by a simple, closed, piecewise-smooth boundary curve $C$. If the orientations of $S$ and $C$ are consistent, meaning that if you traverse $C$ in the direction given by the orientation, then $S$ will always be on your left side, then Stokes' theorem states that:

${\oint}_{\ufffd}\mathbf{\ufffd}\cdot \ufffd\mathbf{\ufffd}={\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}$

where $\oint_C$ represents the line integral around the boundary curve $C$, $\mathbf{F} \cdot d\mathbf{r}$ represents the dot product between the vector field $\mathbf{F}$ and the differential displacement vector $d\mathbf{r}$ along $C$, $\iint_S$ represents the surface integral over the surface $S$, $\nabla \times \mathbf{F}$ represents the curl of the vector field $\mathbf{F}$, and $\cdot$ denotes the dot product between vectors.

To prove Stokes' theorem, we start by considering a small piece of the surface $S$, denoted as $d\mathbf{S}$. We can parameterize this small piece of surface using two parameters, say $u$ and $v$, such that $\mathbf{r}(u, v)$ gives the position vector of a point on the surface. The unit normal vector to the surface $d\mathbf{S}$ at each point can be represented as $\mathbf{n}(u, v)$. Then, the vector differential $d\mathbf{S}$ is given by the cross product of the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$:

$\ufffd\mathbf{\ufffd}=\frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd}\times \frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd}\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd$

Next, we consider a small piece of the boundary curve $C$, denoted as $d\mathbf{r}$. We parameterize this curve using a parameter $t$, such that $\mathbf{r}(t)$ gives the position vector of a point on the curve. The tangent vector to the curve $d\mathbf{r}$ at each point can be represented as $\frac{d\mathbf{r}}{dt}$. Then, the vector differential $d\mathbf{r}$ is given by:

$\ufffd\mathbf{\ufffd}=\frac{\ufffd\mathbf{\ufffd}}{\ufffd\ufffd}\text{\hspace{0.17em}}\ufffd\ufffd$

Now, let's compute the line integral $\oint_C \mathbf{F} \cdot d\mathbf{r}$ by substituting the expressions for $d\mathbf{r}$ and $d\mathbf{S}$:

${\oint}_{\ufffd}\mathbf{\ufffd}\cdot \ufffd\mathbf{\ufffd}={\oint}_{\ufffd}\mathbf{\ufffd}\cdot \left(\frac{\ufffd\mathbf{\ufffd}}{\ufffd\ufffd}\text{\hspace{0.17em}}\ufffd\ufffd\right)$

Using the chain rule, we can express $\frac{d\mathbf{r}}{dt}$ in terms of $u$ and $v$, and rewrite the line integral as:

${\oint}_{\ufffd}\mathbf{\ufffd}\cdot \ufffd\mathbf{\ufffd}={\oint}_{\ufffd}\mathbf{\ufffd}\cdot \left(\frac{\ufffd\mathbf{\ufffd}}{\ufffd\ufffd}\text{\hspace{0.17em}}\ufffd\ufffd\right)={\oint}_{\ufffd}\mathbf{\ufffd}\cdot (\frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd}\times \frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd$

Now, let's compute the surface integral $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ by substituting the expression for $d\mathbf{S}$:

${\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}={\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot (\frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd}\times \frac{\mathrm{\partial}\mathbf{\ufffd}}{\mathrm{\partial}\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd$

Using the vector identity $\nabla \times \mathbf{F} \cdot \mathbf{n} = \mathbf{F} \cdot (\nabla \times \mathbf{n}) - \mathbf{n} \cdot (\nabla \times \mathbf{F})$, where $\mathbf{n}$ is the unit normal vector to the surface, we can rewrite the surface integral as:

${\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}={\iint}_{\ufffd}\mathbf{\ufffd}\cdot (\mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd-{\iint}_{\ufffd}\mathbf{\ufffd}\cdot (\mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd$

The term $\mathbf{n} \cdot (\nabla \times \mathbf{F})$ represents the dot product between the unit normal vector $\mathbf{n}$ and the curl of the vector field $\mathbf{F}$. Since $S$ is a smooth surface, the unit normal vector $\mathbf{n}$ is continuous across $S$, which implies that $\mathbf{n}$ is the same on both sides of $S$. Hence, $\mathbf{n} \cdot (\nabla \times \mathbf{F})$ is zero, and the surface integral simplifies to:

${\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}={\iint}_{\ufffd}\mathbf{\ufffd}\cdot (\mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd$

Now, by applying the divergence theorem to the surface integral on the right-hand side, we obtain:

${\iint}_{\ufffd}\mathbf{\ufffd}\cdot (\mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd\text{\hspace{0.17em}}\ufffd\ufffd={\iiint}_{\ufffd}\mathrm{\nabla}\cdot (\mathbf{\ufffd}\times \mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd$

Since the unit normal vector $\mathbf{n}$ is continuous across $S$, its curl $\nabla \times \mathbf{n}$ represents twice the mean curvature vector of $S$, which is zero for a flat surface. Therefore, $\mathbf{F} \cdot (\nabla \times \mathbf{n}) = 0$, and the volume integral simplifies to zero:

${\iiint}_{\ufffd}\mathrm{\nabla}\cdot (\mathbf{\ufffd}\times \mathrm{\nabla}\times \mathbf{\ufffd})\text{\hspace{0.17em}}\ufffd\ufffd=0$

As a result, the surface integral becomes zero:

${\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}=0$

Hence, we have:

${\oint}_{\ufffd}\mathbf{\ufffd}\cdot \ufffd\mathbf{\ufffd}={\iint}_{\ufffd}(\mathrm{\nabla}\times \mathbf{\ufffd})\cdot \ufffd\mathbf{\ufffd}=0$

This completes the proof of Stokes' theorem.

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