Section 6.8 Orientation of Surface 

6.8 Orientation of Surface 

When we do the line integration or curve integration, we integrate the curve using the counterclockwise direction as the positive orientation.

For the surface integration, we need to define what is a positive orientation of a surface. In general, for a closed surface, for example sphere, we use outward as the positive orientation. For [latex]z=f(x,y)[/latex], we use “facing up” as the positive orientation, i.e. the [latex]z[/latex]-component is positive. We need to find a way to obtain the positive orientation of surface. For a smooth surface, we can find tangent plane at every point of the surface. Then we can find normal vectors of the tangent plane with exactly opposite direction of each other. 

Suppose [latex]z=f(x,y)[/latex] is a surface, then [latex]\overrightarrow{r(x,y)}=\lt x,y,z=f(x,y)\gt[/latex] is the parametric vector function. The normal vector can be obtained by [latex]\overrightarrow{r_{x}}\times\overrightarrow{r_{y}}[/latex] or [latex]\overrightarrow{r_{y}}\times\overrightarrow{r_{x}}[/latex]. In general, a surface in 3-dimensional space, if we close our 4 fingers with counterclockwise orientation on the boundary of the surface, then the thumb gives the positive orientation. This makes sense with cross product, and the right hand rule for 3-dimensional space. 

There is one special surface that cannot define a positive orientation: Möbius strip. We do not consider this kind of surface in this course. 

Definition: Surface Integrals of Vector Fields

Let [latex]\overrightarrow{F}[/latex] be a continuous vector field with a domain that contains oriented surface [latex]S[/latex] with unit normal vector [latex]\overrightarrow{N}[/latex]. The surface integral of [latex]\overrightarrow{F}[/latex] over [latex]S[/latex] is 

\[\int\int_{S}\overrightarrow{F}\cdot dS =\int\int_{S}\overrightarrow{F}\cdot\overrightarrow{N}dS\]

\[=\int\int_{D}\overrightarrow{F}\cdot\overrightarrow{r_{u}}\times\overrightarrow{r_{v}}dA\]

Where [latex]\overrightarrow{r(u,v)}[/latex] is the vector function of the surface and [latex]D[/latex] is the parameter domain. This integral is also called the flux of [latex]\overrightarrow{F}[/latex] across [latex]S[/latex]. 

Example 1: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the parallelogram [latex]\lt u-v,1+v+2u,u+v\gt[/latex] with [latex]0\leq u\leq1[/latex] and [latex]0\leq v\leq2[/latex], and [latex]\overrightarrow{F}=\lt 2x,y,z\gt[/latex]. 

Exercise 1: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the parallelogram [latex]\lt 1-u+v,2v-u,u-v\gt[/latex] with [latex]0\leq u\leq2[/latex] and [latex]0\leq v\leq1[/latex], and [latex]\overrightarrow{F}=\lt x,-y,2z\gt[/latex]. 

Example 2: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the vector function [latex]\lt v,u\text{cos}(v),u\text{sin}(v)\gt[/latex] with [latex]0\leq u\leq1[/latex] and [latex]0\leq v\leq\frac{\pi}{2}[/latex], and [latex]\overrightarrow{F}=\lt x,-z,y\gt[/latex].

Exercise 2: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the vector function [latex]\lt v\text{sin}(u),v\text{cos}(u),u\gt[/latex] with [latex]0\leq u\leq\frac{\pi}{2}[/latex] and [latex]0\leq v\leq1[/latex], and [latex]\overrightarrow{F}=\lt y,-x,z\gt[/latex].

For a surface [latex]z=g(x,y)[/latex], we can define a parametric vector function using [latex]\overrightarrow{r(x,y)}=\lt x,y,g(x,y)\gt[/latex], hence we have [latex]\overrightarrow{r_{x}}\times\overrightarrow{r_{y}}=\lt -g_{x},-g_{y},1\gt[/latex] and the flux of [latex]\overrightarrow{F}[/latex] over [latex]S[/latex] is 

\[\int\int_{S}\overrightarrow{F}\cdot dS=\int\int_{D}\overrightarrow{F}\cdot\lt -g_{x},-g_{y},1\gt dA.\]

Example 3: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]\overrightarrow{F}=x\overrightarrow{i}+2y\overrightarrow{j}-3z\overrightarrow{k}[/latex], and [latex]S[/latex] is that part of plane [latex]x-2y+z=6[/latex] that lies above unit square [latex]0\le x\le1,0\le y\le1[/latex].

Exercise 3: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]\overrightarrow{F}=x\overrightarrow{i}-y\overrightarrow{j}+z\overrightarrow{k}[/latex], and [latex]S[/latex] is that part of plane [latex]3x-y+z=8[/latex] that lies above unit square [latex]0\le x\le1,0\le y\le1[/latex].

Example 4: Find the flux using negative (downward) orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the paraboloid [latex]z=4-x^{2}-y^{2}[/latex] from [latex]z=0[/latex] to [latex]z=1[/latex], and [latex]\overrightarrow{F}=\lt y,-x,z\gt[/latex]. 

Exercise 4: Find the flux using negative (downward) orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the paraboloid [latex]z=x^{2}+y^{2}[/latex] from [latex]z=0[/latex] to [latex]z=1[/latex], and [latex]\overrightarrow{F}=\lt x,y,z\gt[/latex]. 

Example 5: Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the boundary of [latex]0\leq x\leq\sqrt{1-z^{2}}[/latex] from [latex]y=0[/latex] to [latex]y=3[/latex], and [latex]\overrightarrow{F}=\lt x,y,z\gt[/latex]. 

Group work:

1. Find the flux using positive orientation. [latex]\int\int_{S}\overrightarrow{F}\cdot dS[/latex] where [latex]S[/latex] is the boundary of tetrahedron bounded by three vertices: [latex](0,0,0)[/latex], [latex](1,0,0)[/latex], [latex](0,1,0)[/latex] and [latex](0,0,1)[/latex], and [latex]\overrightarrow{F}=\lt y,z-y,x\gt[/latex].