Introduction
In this part we’ll cover surface integrals over vector fields as well as two very important theorems.
Surface integrals over vector fields
Definition 1 (Flux) Let S S S r ⃗ : D → R 2 \vec{r}: D \rarr \mathbb{R}^2 r : D → R 2 F : R 3 → R 3 F: \mathbb{R}^3 \rarr \mathbb{R}^3 F : R 3 → R 3
∬ S F ⃗ ⋅ d S = ∬ D F ⃗ ( r ⃗ ( u , v ) ) ⋅ r u ⃗ × r v ⃗ d u d v \iint_S \vec{F} \cdot dS = \iint_D \vec{F}(\vec{r}(u, v)) \cdot \vec{r_u} \times \vec{r_v}\ du\ dv ∬ S F ⋅ d S = ∬ D F ( r ( u , v )) ⋅ r u × r v d u d v We call this the flux of F ⃗ \vec{F} F S S S
Example 1 Given, F ⃗ = ⟨ z , y , x ⟩ \vec{F} = \langle z, y, x \rangle F = ⟨ z , y , x ⟩
So, we have the surface:
S = { ( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 } S = \{(x, y, z) | x^2 + y^2 + z^2 \leq 1 \} S = {( x , y , z ) ∣ x 2 + y 2 + z 2 ≤ 1 } As always, good idea to change into polar coordinates, so a good parameterization is:
r ⃗ ( ϕ , θ ) = ⟨ s i n ( ϕ ) c o s ( θ ) , s i n ( ϕ ) s i n ( θ ) , c o s ( ϕ ) ⟩ ∣ 0 ≤ ϕ ≤ π , 0 ≤ θ ≤ 2 π \vec{r}(\phi, \theta) = \langle sin(\phi)\ cos(\theta), sin(\phi)\ sin(\theta), cos(\phi) \rangle \ | \ 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi r ( ϕ , θ ) = ⟨ s in ( ϕ ) cos ( θ ) , s in ( ϕ ) s in ( θ ) , cos ( ϕ )⟩ ∣ 0 ≤ ϕ ≤ π , 0 ≤ θ ≤ 2 π NB: This computation is just tedious (and we’ll see how we can reduce this shortly)
r u ⃗ × r v ⃗ = ⟨ s i n 2 ( ϕ ) c o s ( θ ) , s i n 2 ( ϕ ) s i n ( θ ) , s i n ( ϕ ) c o s ( ϕ ) ⟩ \vec{r_u} \times \vec{r_v} = \langle sin^2(\phi)\ cos(\theta), sin^2(\phi)\ sin(\theta), sin(\phi)\ cos(\phi) \rangle r u × r v = ⟨ s i n 2 ( ϕ ) cos ( θ ) , s i n 2 ( ϕ ) s in ( θ ) , s in ( ϕ ) cos ( ϕ )⟩ Now, one can solve this integral by taking the dot product and applying our definition:
∫ 0 2 π ∫ 0 π ⟨ c o s ( ϕ ) , s i n ( ϕ ) s i n ( θ ) , s i n ( ϕ ) c o s ( θ ) ⟩ ⋅ ⟨ s i n 2 ( ϕ ) c o s ( θ ) , s i n 2 ( ϕ ) s i n ( θ ) , s i n ( ϕ ) c o s ( ϕ ) ⟩ \int_0^{2\pi} \int_0^{\pi} \langle cos(\phi), sin(\phi)\ sin(\theta), sin(\phi)\ cos(\theta) \rangle \cdot \langle sin^2(\phi)\ cos(\theta), sin^2(\phi)\ sin(\theta), sin(\phi)\ cos(\phi) \rangle ∫ 0 2 π ∫ 0 π ⟨ cos ( ϕ ) , s in ( ϕ ) s in ( θ ) , s in ( ϕ ) cos ( θ )⟩ ⋅ ⟨ s i n 2 ( ϕ ) cos ( θ ) , s i n 2 ( ϕ ) s in ( θ ) , s in ( ϕ ) cos ( ϕ )⟩ Let’s see now how we can do the same example, with fewer computations, first some theory.
Orientation
For surfaces, we say that positive orientation is “pointing outwards”. For hollow surfaces this is always the case. If our parameterization is given by a single function, then we always have the correct choice.
However, imagine half the unit sphere with a filled base. The base with parameterization: D = { ( r , θ ) ∣ 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π } D = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} D = {( r , θ ) ∣0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π }
Definition 2 (Divergence) For a vector field, F ⃗ : R 3 → R 3 \vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3 F : R 3 → R 3 divergence is:
d i v F ⃗ = ∂ P ∂ x + ∂ Q ∂ y + ∂ R ∂ z div\ \vec{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z} d i v F = ∂ x ∂ P + ∂ y ∂ Q + ∂ z ∂ R Theorem 1 (The Divergence Theorem) Let S S S E E E ∂ E = S \partial E = S ∂ E = S
Then:
∬ S F ⃗ ⋅ d S = ∭ E d i v F ⃗ d V \iint_S \vec{F} \cdot dS = \iiint_E div\ \vec{F}\ dV ∬ S F ⋅ d S = ∭ E d i v F d V Equivalently:
∬ ∂ E F ⃗ ⋅ d S = ∭ E d i v F ⃗ d V \iint_{\partial E} \vec{F} \cdot dS = \iiint_E div\ \vec{F}\ dV ∬ ∂ E F ⋅ d S = ∭ E d i v F d V Now let’s take the sample example.
Example 2 Find the flux of F ⃗ = ⟨ z , y , x \vec{F} = \langle z, y, x F = ⟨ z , y , x
d i v F ⃗ = ∂ P ∂ x + ∂ Q ∂ y + ∂ R ∂ z = 0 + 0 + 1 = 1 div\ \vec{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z} = 0 + 0 + 1 = 1 d i v F = ∂ x ∂ P + ∂ y ∂ Q + ∂ z ∂ R = 0 + 0 + 1 = 1 ∬ S F ⃗ ⋅ d S = ∭ E 1 d V \iint_S \vec{F} \cdot dS = \iiint_E 1 dV ∬ S F ⋅ d S = ∭ E 1 d V We could now just transform to polar coordinates, but, this is the volume of the unit sphere, which is just:
4 π 3 \boxed{\dfrac{4\pi}{3}} 3 4 π Stokes’ Theorem
So far we have seen Green’s Theorem as well as The Divergence theorem. But what is the reason that Green’s Theorem only works in R 3 \mathbb{R}^3 R 3
Well, Stokes Theorem is essentially Green’s Theorem but in R 3 \mathbb{R}^3 R 3
Definition 3 (Curl) For F ⃗ : R 3 → R 3 \vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3 F : R 3 → R 3 curl as:
NB: it is also called rot and sometimes denoted as ∇ × F ⃗ \nabla \times \vec{F} ∇ × F
c u r l F ⃗ = ⟨ ∂ R ∂ y − ∂ Q ∂ z , ∂ P ∂ z − ∂ R ∂ x , ∂ Q ∂ x − ∂ P ∂ y ⟩ curl\ \vec{F} = \langle \dfrac{\partial R}{\partial y} - \dfrac{\partial Q}{\partial z}, \dfrac{\partial P}{\partial z} - \dfrac{\partial R}{\partial x}, \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \rangle c u r l F = ⟨ ∂ y ∂ R − ∂ z ∂ Q , ∂ z ∂ P − ∂ x ∂ R , ∂ x ∂ Q − ∂ y ∂ P ⟩ Theorem 2 (Stokes’ Theorem) If S ⊂ R 3 S \subset \mathbb{R}^3 S ⊂ R 3 R 3 \mathbb{R}^3 R 3
∫ C F ⃗ ⋅ d r = ∬ S c u r l F d S \int_C \vec{F} \cdot dr = \iint_S curl\ F\ dS ∫ C F ⋅ d r = ∬ S c u r l F d S Example 3 Compute ∫ C F ⋅ d r \int_C F \cdot dr ∫ C F ⋅ d r F ⃗ = ⟨ − y 2 , x , z 2 ⟩ \vec{F} = \langle -y^2, x, z^2 \rangle F = ⟨ − y 2 , x , z 2 ⟩ C C C y + z = 2 y + z = 2 y + z = 2 x 2 + y 1 = 1 x^2 + y^1 = 1 x 2 + y 1 = 1
Let’s find the curl:
c u r l F ⃗ = ⟨ 0 , 0 , 1 + 2 y ⟩ curl\ \vec{F} = \langle 0, 0, 1 + 2y \rangle c u r l F = ⟨ 0 , 0 , 1 + 2 y ⟩ We need to find the surface, the intersection of y + z = 2 y + z = 2 y + z = 2 x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1
∬ S c u r l F ⃗ d S \iint_S curl\ \vec{F}\ dS ∬ S c u r l F d S Is just the flux through this region, let’s express this in polar coordinates
D = { ( r , θ ) ∣ 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π } D = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \} D = {( r , θ ) ∣0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π } Meaning:
∬ S c u r l F ⃗ d S = ∬ D c u r l F ⃗ ⋅ r u ⃗ × r v ⃗ d A \iint_S curl\ \vec{F}\ dS = \iint_D curl\ \vec{F} \cdot \vec{r_u} \times \vec{r_v}\ dA ∬ S c u r l F d S = ∬ D c u r l F ⋅ r u × r v d A Let’s find a parameterization of the surface:
r ⃗ ( u , v ) = ⟨ u , v , 2 − v ⟩ \vec{r}(u, v) = \langle u, v, 2 - v \rangle r ( u , v ) = ⟨ u , v , 2 − v ⟩ r u ⃗ ( u , v ) = ⟨ 1 , 0 , 0 ⟩ \vec{r_u}(u, v) = \langle 1, 0, 0 \rangle r u ( u , v ) = ⟨ 1 , 0 , 0 ⟩ r v ⃗ ( u , v ) = ⟨ 0 , 1 , − 1 ⟩ \vec{r_v}(u, v) = \langle 0, 1, -1 \rangle r v ( u , v ) = ⟨ 0 , 1 , − 1 ⟩ r u ⃗ × r v ⃗ = ⟨ 0 , 1 , 1 ⟩ \vec{r_u} \times \vec{r_v} = \langle 0, 1, 1 \rangle r u × r v = ⟨ 0 , 1 , 1 ⟩ ∬ D c u r l F ⃗ ⋅ r u ⃗ × r v ⃗ d A = ∬ D ⟨ 0 , 0 , 1 + 2 y ⟩ ⋅ ⟨ 0 , 1 , 1 ⟩ d A \iint_D curl\ \vec{F} \cdot \vec{r_u} \times \vec{r_v}\ dA = \iint_D \langle 0, 0, 1 + 2y \rangle \cdot \langle 0, 1, 1 \rangle\ dA ∬ D c u r l F ⋅ r u × r v d A = ∬ D ⟨ 0 , 0 , 1 + 2 y ⟩ ⋅ ⟨ 0 , 1 , 1 ⟩ d A ∬ D 1 + 2 y d A \iint_D 1 + 2y\ dA ∬ D 1 + 2 y d A ∫ 0 2 π ∫ 0 1 ( 1 + 2 r s i n ( θ ) ) r d r d θ \int_0^{2\pi} \int_0^1 (1 + 2r sin(\theta))r\ dr\ d\theta ∫ 0 2 π ∫ 0 1 ( 1 + 2 r s in ( θ )) r d r d θ ∫ 0 2 π ∫ 0 1 r + 2 r 2 s i n ( θ ) d r d θ \int_0^{2\pi} \int_0^1 r + 2r^2 sin(\theta)\ dr\ d\theta ∫ 0 2 π ∫ 0 1 r + 2 r 2 s in ( θ ) d r d θ ∫ 0 1 ∫ 0 2 π r + 2 r 2 s i n ( θ ) d θ d r \int_0^1 \int_0^{2\pi} r + 2r^2 sin(\theta)\ d\theta\ dr ∫ 0 1 ∫ 0 2 π r + 2 r 2 s in ( θ ) d θ d r s i n ( θ ) sin(\theta) s in ( θ ) 2 π 2\pi 2 π 0 0 0
∫ 0 1 2 π r d r = π \int_0^1 2\pi r\ dr = \boxed{\pi} ∫ 0 1 2 π r d r = π Example 4 Compute ∬ S c u r l F ⃗ d S \iint_S curl\ \vec{F}\ dS ∬ S c u r l F d S F ⃗ = ⟨ x z , y z , x y ⟩ \vec{F} = \langle xz, yz, xy \rangle F = ⟨ x z , y z , x y ⟩ S S S ( 0 , 0 , 0 ) (0, 0, 0) ( 0 , 0 , 0 ) x 2 + y 2 ≤ 1 x^2 + y^2 \leq 1 x 2 + y 2 ≤ 1 x y xy x y
Let’s actually use Stokes theorem, “backwards”. C C C
{ x 2 + y 2 = 1 x 2 + y 2 + z 2 = 4 \begin{cases}
x^2 + y^2 = 1 \newline
x^2 + y^2 + z^2 = 4 \newline
\end{cases} { x 2 + y 2 = 1 x 2 + y 2 + z 2 = 4 Which means that z = 3 z = \sqrt{3} z = 3
We can parameterize this curve as:
r ⃗ ( t ) = ⟨ c o s ( t ) , s i n ( t ) , 3 ⟩ ∣ 0 ≤ t ≤ 2 π \vec{r}(t) = \langle cos(t), sin(t), \sqrt{3} \rangle \ | \ 0 \leq t \leq 2\pi r ( t ) = ⟨ cos ( t ) , s in ( t ) , 3 ⟩ ∣ 0 ≤ t ≤ 2 π r ′ ⃗ ( t ) = ⟨ − s i n ( t ) , c o s ( t ) , 0 ⟩ \vec{r^\prime}(t) = \langle -sin(t), cos(t), 0 \rangle r ′ ( t ) = ⟨ − s in ( t ) , cos ( t ) , 0 ⟩ ∬ S c u r l F ⃗ d S = ∫ C F ⃗ ⋅ d r \iint_S curl\ \vec{F}\ dS = \int_C \vec{F} \cdot dr ∬ S c u r l F d S = ∫ C F ⋅ d r ∫ 0 2 π ⟨ 3 c o s ( t ) , 3 s i n ( t ) , c o s ( t ) s i n ( t ) ⟩ ⋅ ⟨ − s i n ( t ) , c o s ( t ) , 0 ⟩ \int_0^{2\pi} \langle \sqrt{3} cos(t), \sqrt{3} sin(t), cos(t)\ sin(t) \rangle \cdot \langle -sin(t), cos(t), 0 \rangle ∫ 0 2 π ⟨ 3 cos ( t ) , 3 s in ( t ) , cos ( t ) s in ( t )⟩ ⋅ ⟨ − s in ( t ) , cos ( t ) , 0 ⟩ ∫ 0 2 π − 3 c o s ( t ) s i n ( t ) + 3 s i n ( t ) c o s ( t ) \int_0^{2\pi} -\sqrt{3} cos(t)\ sin(t) + \sqrt{3} sin(t)\ cos(t) ∫ 0 2 π − 3 cos ( t ) s in ( t ) + 3 s in ( t ) cos ( t ) ∫ 0 2 π 0 = 0 \int_0^{2\pi} 0 = \boxed{0} ∫ 0 2 π 0 = 0