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The **Kelvin–Stokes theorem**,^{[1]}^{[2]}^{[3]}^{[4]}^{[5]}
also known as the **curl theorem**,^{[6]} is a theorem in vector calculus on **R**^{3}. Given a vector field, the theorem relates the integral of the curl of the vector field over some surface, to the line integral of the vector field around the boundary of the surface. The Kelvin–Stokes theorem is a special case of the “generalized Stokes' theorem.”^{[7]}^{[8]} In particular, the vector field on **R**^{3} can be considered as a 1-form in which case curl is the exterior derivative.

## Theorem

Let γ : [*a*, *b*] → **R**^{2} be a Piecewise smooth Jordan plane curve, that bounds the domain *D* ⊂ **R**^{2}.^{[note 1]} Suppose ψ : *D* → **R**^{3} is smooth, with *S* := ψ[*D*]
^{[note 2]}
, and Γ is the space curve defined by Γ(*t*) = ψ(γ(*t*)).^{[note 3]} If **F** a smooth vector field on **R**^{3}, then^{[1]}^{[2]}^{[3]}

- $\oint _{\Gamma }\mathbf {F} \,d\Gamma =\iint _{S}\nabla \times \mathbf {F} \,dS$

## Proof

The proof of the Theorem consists of 4 steps.^{[2]}^{[3]}^{[note 4]} We assume that the Green's theorem is known, so what is of concern is "how to boil down the three-dimensional complicated problem (Kelvin–Stokes theorem) to a two-dimensional rudimentary problem (Green's theorem)". In an ordinary, mathematicians use the differential form, especially "pull-back^{[note 4]} of differential form" is very powerful tool for this situation, however, learning differential form needs too much background knowledge. So, the proof below does not require background information on differential form, and may be helpful for understanding the notion of differential form.

### First step of the proof (defining the pullback)

Define

- ${\mathbf {P} }(u,v)=(P_{1}(u,v),P_{2}(u,v))$

so that **P** is the pull-back^{[note 4]} of **F**, and that **P**(*u*, *v*) is **R**^{2}-valued function, depends on two parameter *u*, *v*. In order to do so we define *P*_{1} and *P*_{2} as follows.

- $P_{1}(u,v)=\left\langle {\mathbf {F} }(\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle ,\qquad P_{2}(u,v)=\left\langle {\mathbf {F} }(\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle$

Where, $\langle \ |\ \rangle$ is the normal inner product of **R**^{3} and
hereinafter, $\langle \ |A|\ \rangle$ stands for bilinear form according to matrix **A** ^{[note 5]}
.^{[note 6]}

### Second step of the proof (first equation)

According to the definition of line integral,

- ${\begin{aligned}\oint _{\Gamma }\mathbf {F} d\Gamma &=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}{\frac {d\Gamma }{dt}}(t)\right\rangle \,dt\\&=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}{\frac {d(\psi \circ \gamma )}{dt}}(t)\right\rangle \,dt\\&=\int _{a}^{b}\left\langle (\mathbf {F} \circ \psi (t)){\bigg |}(J\psi )_{\gamma (t)}\cdot {\frac {d\gamma }{dt}}(t)\right\rangle \,dt\end{aligned}}$

where, *J*ψ stands for the Jacobian matrix of ψ. Hence,^{[note 5]}^{[note 6]}

- ${\begin{aligned}\left\langle ({\mathbf {F} }\circ \Gamma (t)){\bigg |}(J\psi )_{\gamma (t)}{\frac {d\gamma }{dt}}(t)\right\rangle &=\left\langle ({\mathbf {F} }\circ \Gamma (t)){\bigg |}(J\psi )_{\gamma (t)}{\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle ({}^{t}{\mathbf {F} }\circ \Gamma (t))\cdot (J\psi )_{\gamma (t)}\ {\bigg |}\ {\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle \left(\left\langle ({\mathbf {F} }(\psi (\gamma (t)))){\bigg |}{\frac {\partial \psi }{\partial u}}(\gamma (t))\right\rangle ,\left\langle ({\mathbf {F} }(\psi (\gamma (t)))){\bigg |}{\frac {\partial \psi }{\partial v}}(\gamma (t))\right\rangle \right){\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle (P_{1}(u,v),P_{2}(u,v)){\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \\&=\left\langle {\mathbf {P} }(u,v)\ {\bigg |}{\frac {d\gamma }{dt}}(t)\right\rangle \end{aligned}}$

So, we obtain following equation

- $\oint _{\Gamma }{\mathbf {F} }d\Gamma =\oint _{\gamma }{\mathbf {P} }d\gamma$

### Third step of the proof (second equation)

First, calculate the partial derivatives, using Leibniz rule of inner product

- ${\begin{aligned}{\frac {\partial P_{1}}{\partial v}}&=\left\langle {\frac {\partial ({\mathbf {F} }\circ \psi )}{\partial v}}{\bigg |}{\frac {\partial \psi }{\partial u}}\right\rangle +\left\langle {\mathbf {F} }\circ \psi {\bigg |}{\frac {\partial ^{2}\psi }{\partial v\partial u}}\right\rangle \\{\frac {\partial P_{2}}{\partial u}}&=\left\langle {\frac {\partial ({\mathbf {F} }\circ \psi )}{\partial u}}{\bigg |}{\frac {\partial \psi }{\partial v}}\right\rangle +\left\langle {\mathbf {F} }\circ \psi {\bigg |}{\frac {\partial ^{2}\psi }{\partial u\partial v}}\right\rangle \end{aligned}}$

So,^{[note 5]} ^{[note 6]}
^{[note 7]}

- ${\begin{aligned}\iint _{S}(\nabla \times \mathbf {F} )\,dS&=\iint _{D}\left\langle (\nabla \times \mathbf {F} )(\psi (u,v)){\bigg |}{\frac {\partial \psi }{\partial u}}(u,v)\times {\frac {\partial \psi }{\partial v}}(u,v)\right\rangle \,du\,dv\\&=\iint _{D}\det \left[(\nabla \times \mathbf {F} )(\psi (u,v))\quad {\frac {\partial \psi }{\partial u}}(u,v)\quad {\frac {\partial \psi }{\partial v}}(u,v)\right]\,du\,dv&&{\text{ scalar triple product}}\end{aligned}}$