Surface integral of the second kind

4.6 Surface integral of the second kind

Where to do the integral: on the oriented surface
$Σ = ∖left{∖lef t.∖right|∖right}$
For which to do the integral: vector field $F$ .

Definition 4.6.1 Surface $Σ$ with a continuous unit norm vector field $n : Σ \to R^{3}$ is a oriented (orientable) surface $(Σ, n)$ .

Example 4.6.1 See Figure 4.6.

Sphere is orientable. $Σ : x^{2} + y^{2} + z^{2} = R^{2}$ , then
$1 n(x,y,z ) = --(x, y,z)T R$
Tyre surface is orientable.
Möbius strip is non-orientable.

Figure 4.6: Orientable and non-orientable surface

Definition 4.6.2 Assuming $(Σ, n)$ is a oriented surface, $F : Σ \to R^{3}$ is continuous, the surface integral of the second kind of $F$ along $(Σ, n)$ is defined as $\int_{Σ} ⟨ F, n ⟩ d σ$ .

Assuming

-------1-------- n(x) = √ --2----2-----2, F = A + B + C

then we have

AX + BY + CZ 1 ∂x ∂x ⟨F, n⟩ = -√--2-----2----2- = √---2----2-----2∖left⟨F, ---× --∖right ⟩ A + B + C A + B + C ∂u ∂v

At this time, if $(A, B, C) = \frac{\partial x}{\partial u} \times \frac{\partial x}{\partial v}$ , i.e.

∂x ∂x ∂ (y,z) ∂(z,x) ∂(x,y ) ---× ---= det-------i + det -------j + det-------k ∂u ∂v ◟--∂◝(u◜,-v)◞ ◟---∂◝(◜u,v)◞ ◟---∂◝(◜u,v-)◞ A B C

we can prove that

√ ---------- √ -------------- dσ = EG − F 2dudv = A2 + B2 + C2dudv

Assuming $D : {(u, v)}$ Hence

∫ ∫ ∫ ⟨F, n⟩dσ = ∖left⟨F, ∂x-× ∂x-∖right⟩dudv = det∖left(F, ∂x-, ∂x-∖right)dudv Σ D ∂u ∂v D ∂u ∂v

The geometric (physical) meaning of the formula above. See Figure 4.7, $det (F, \frac{\partial x}{\partial u}, \frac{\partial x}{\partial v})$ represents the volume of infinitesimal flow.

Note Here, the order of $u, v$ represents the direction of $Σ$ . See Figure 4.8, when we use spherical coordinate to describe a sphere, we use $(θ, ϕ)$ if the norm vector is outer norm vector, $(ϕ, θ)$ if the norm vector is inner norm vector.

Figure 4.7: Physical meaning of the integral

Figure 4.8: The order of parametric variable

Define

∂ (x, y) dx ∧ dy = det -------dudv ∂ (u, v)

then we could rewrite the integral above as

∫ ∫ ∫ ⟨F,n ⟩dσ = Xdy ∧ dz + Y dz ∧ dx + Zdx ∧ dy = ω Σ Σ Σ

where $ω = X d y \land d z + Y d z \land d x + Z d x \land d y$ is called the form of second-order derivative. If the direction of projection of the norm vector of the surface is the same as the direction of the norm vector of $x O y$ plane, then $d x \land d y = d x d y$ , otherwise $d x \land d y = - d x d y$ . The geometric meaning of this form is shown in Figure 4.9.

Figure 4.9: Geometric meaning of the integral

Properties of the integral.

Linearity of $F$ (or $ω$ ).
Additivity of $Σ$ .

Example 4.6.2 Assuming $Σ : z = x^{2} + y^{2} (x^{2} + y^{2} \leq 1)$ where $z$ component of the norm vector points downward ( $- z$ direction), seek $\int_{Σ} x d y \land d z$ .

Method 1. Since $i \times j$ get $k$ , so $(x, y)$ corresponds with norm vector pointing $+ z$ direction, and $(y, x)$ corresponds with $- z$ direction. The parametric representative of $Σ$ is

x =, D : x2 + y2 ≤ 1

and the vector field

F ==

Thus

∫ ∫ ∂x ∂x ∫ ∫ ω = det ∖lef t(F,---, --∖right ) = 2x2dxdy = (x2 + y2)dxdy Σ D ∂y ∂x D D

Method 2.

∫ ∫ ∫ ∫ 2 2 2 2 I = Σxdy ∧ d(x + y ) = Σ xdy ∧ (2xdx + 2ydy ) = Σ 2x dy ∧ dx = D 2x dxdy ◟-----◝◜-----◞ ◟----◝◜---◞ Surface integral Multiple integral

Example 4.6.3 Assuming $Σ : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1$ where the norm vector points outward. Seek $\int_{Σ} x d y \land d z + y d z \land d x + z d x \land d y$ .

Let

\text{[math]}

i.e.

(x,y, z)→ ∖left(x-, y-, z∖right) = (˜x,y˜, ˜z) ◟--◝◜--◞ ◟-----a--b--c-◝◜--------------◞ Ellipsoid Sphere

For the sphere, $(θ, ϕ)$ corresponds with $+ r$ direction. If

∂ (x˜, ˜y, ˜z) det--------- > 0 ∂ (x, y,z)

then the transformation is a direction-preserving transformation, meaning for the ellipsoid surface, $(θ, ϕ)$ also corresponds with $+ r$ direction. Therefore $\begin{aligned} \int_{Σ} ω & = \int_{\tilde{Σ}} a \tilde{x} d (b \tilde{y}) \land d (c \tilde{z}) + b \tilde{y} d (c \tilde{z}) \land d (a \tilde{x}) + c \tilde{z} d (a \tilde{x}) \land d (b \tilde{y}) \\ = a b c \int_{\tilde{Σ}} \tilde{x} d \tilde{y} \land d \tilde{z} + \tilde{y} d \tilde{z} \land d \tilde{x} + \tilde{z} d \tilde{x} \land d \tilde{y} \\ = a b c \int_{\tilde{Σ}} ⟨ n, n ⟩ d σ = a b c \int_{\tilde{Σ}} d σ = 4 π a b c \end{aligned}$

Example 4.6.4 Assuming $Σ : z = 1 - x^{2} - y^{2} (x^{2} + y^{2} \leq 1)$ where the norm vector points upward. Seek $\int_{Σ} (x^{2} - z) d x \land d y + (z^{2} - y) d z \land d x$ .

Since $d z = - 2 x d x - 2 y d y$ , we have $\begin{aligned} I & = \int_{Σ} (x^{2} - (1 - x^{2} - y^{2})) d x ⟨ d y + [(1 - x^{2} - y^{2})^{2} - y] (- 2 y d y) \land d x \\ = \int_{Σ} [2 x^{2} - y^{2} - 1 + 2 y (1 - x^{2} - y^{2})^{2}] d x \land d y \\ = \int_{D} [2 x^{2} - y^{2} - 1 + \underset{Odd w.r.t. y}{\underset{⏟}{2 y (1 - x^{2} - y^{2})^{2}}}] d x d y \\ = \frac{1}{2} \int_{D} (x^{2} + y^{2} - 2) d x d y \end{aligned}$