Stokes' Theorem

Theorem

Stokes' Theorem is essentially a 33-D version of the curl version of Green's Theorem, which, if you recall, we can write as ∬R∣∇×F⃗∣ dA=∮CF⃗⋅r⃗ ′(t) dt.\iint_R | \nabla \times \vec{F} | \ dA = \oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt.

Instead of a region RR in the plane, we can use a surface SS in R3\mathbb{R}^3. If we make similar assumptions to the ones from Green's Theorem, we get Stokes' Theorem: ∬S(∇×F⃗)⋅n^ dS=∮CF⃗⋅r⃗ ′(t) dt\ans{\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \ dS = \oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt}

Example

Evaluate ∮CF⃗⋅dr⃗\displaystyle\oint_C \vec{F} \cdot d\vec{r} where F⃗=⟨z,−z,x2−y2⟩\vec{F}=⟨z,−z,x^2−y^2⟩ and CC consists of the three line segments bounding the plane z=8−4x−2yz=8−4x−2y in the first octant.

Solution

∮CF⃗⋅dr⃗=∬S(∇×F⃗)⋅n^ dS=∬R(∇×F⃗)⋅(r⃗r×r⃗v) dA∇×F⃗=∣ı^ȷ^k^∂∂x∂∂y∂∂zz−zx2−y2∣=⟨1−2y,1−2x,0⟩r⃗u×r⃗v=∣ı^ȷ^k^10−401−2∣=⟨4,2,1⟩∮CF⃗⋅dr⃗=∫02∫04−2x⟨1−2y,1−2x,0⟩⋅⟨4,2,1⟩ dy dx=∫02∫04−2x6−4x−8y dy dx=−883\begin{aligned} \oint_C \vec{F} \cdot d\vec{r} &= \iint_S (\nabla \times \vec{F}) \cdot \hat{n} \ dS\ &= \iint_R (\nabla \times \vec{F}) \cdot (\vec{r}_r \times \vec{r}_v) \ dA\ \ \nabla \times \vec{F} &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\ z & -z & x^2-y^2 \end{vmatrix}\ \ &= \langle 1-2y,1-2x,0 \rangle\ \ \vec{r}_u \times \vec{r}_v &= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k}\ 1 & 0 & -4\ 0 & 1 & -2 \end{vmatrix}\ \ &= \langle 4,2,1 \rangle\ \ \oint_C \vec{F} \cdot d\vec{r} &= \int_0^2 \int_0^{4-2x} \langle 1-2y,1-2x,0 \rangle \cdot \langle 4,2,1 \rangle\ dy\ dx\ &= \int_0^2 \int_0^{4-2x} 6-4x-8y\ dy\ dx = \ans{-\dfrac{88}{3}} \end{aligned}