Stokes' Theorem
Theorem
Stokes' Theorem is essentially a 3 3 3 -D version of the curl version of Green's Theorem, which, if you recall, we can write as
∬ R ∣ ∇ × F ⃗ ∣ d A = ∮ C F ⃗ ⋅ r ⃗ ′ ( t ) d t . \iint_R | \nabla \times \vec{F} | \ dA = \oint_C \vec{F} \cdot \vec{r}\ '(t) \ dt. ∬ R ∣ ∇ × F ∣ d A = ∮ C F ⋅ r ′ ( t ) d t .
Instead of a region R R R in the plane, we can use a surface S S S in R 3 \mathbb{R}^3 R 3 . If we make similar assumptions to the ones from Green's Theorem, we get Stokes' Theorem:
∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = ∮ C F ⃗ ⋅ r ⃗ ′ ( t ) d t \ans{\iint_S (\nabla \times \vec{F}) \cdot \hat{n} \ dS = \oint_C \vec{F} \cdot \vec{r}\ '(t)\ dt} ∬ S ( ∇ × F ) ⋅ n ^ d S = ∮ C F ⋅ r ′ ( t ) d t
Example
Evaluate ∮ C F ⃗ ⋅ d r ⃗ \displaystyle\oint_C \vec{F} \cdot d\vec{r} ∮ C F ⋅ d r where F ⃗ = ⟨ z , − z , x 2 − y 2 ⟩ \vec{F}=⟨z,−z,x^2−y^2⟩ F = ⟨ z , − z , x 2 − y 2 ⟩ and C C C consists of the three line segments bounding the plane z = 8 − 4 x − 2 y z=8−4x−2y z = 8 − 4 x − 2 y in the first octant.
Solution
∮ C F ⃗ ⋅ d r ⃗ = ∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = ∬ R ( ∇ × F ⃗ ) ⋅ ( r ⃗ r × r ⃗ v ) d A ∇ × F ⃗ = ∣ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z z − z x 2 − y 2 ∣ = ⟨ 1 − 2 y , 1 − 2 x , 0 ⟩ r ⃗ u × r ⃗ v = ∣ ı ^ ȷ ^ k ^ 1 0 − 4 0 1 − 2 ∣ = ⟨ 4 , 2 , 1 ⟩ ∮ C F ⃗ ⋅ d r ⃗ = ∫ 0 2 ∫ 0 4 − 2 x ⟨ 1 − 2 y , 1 − 2 x , 0 ⟩ ⋅ ⟨ 4 , 2 , 1 ⟩ d y d x = ∫ 0 2 ∫ 0 4 − 2 x 6 − 4 x − 8 y d y d x = − 88 3 \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} ∮ C F ⋅ d r ∇ × F r u × r v ∮ C F ⋅ d r = ∬ S ( ∇ × F ) ⋅ n ^ d S = ∬ R ( ∇ × F ) ⋅ ( r r × r v ) d A = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ı ^ ∂ x ∂ z ȷ ^ ∂ y ∂ − z k ^ ∂ z ∂ x 2 − y 2 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ⟨ 1 − 2 y , 1 − 2 x , 0 ⟩ = ∣ ∣ ∣ ∣ ∣ ∣ ı ^ 1 0 ȷ ^ 0 1 k ^ − 4 − 2 ∣ ∣ ∣ ∣ ∣ ∣ = ⟨ 4 , 2 , 1 ⟩ = ∫ 0 2 ∫ 0 4 − 2 x ⟨ 1 − 2 y , 1 − 2 x , 0 ⟩ ⋅ ⟨ 4 , 2 , 1 ⟩ d y d x = ∫ 0 2 ∫ 0 4 − 2 x 6 − 4 x − 8 y d y d x = − 3 8 8