Surface Integral of Flux Type - Example 1

117 days ago by lfahlberg

Calculate the surface integral \iint_S yzdydz+xzdxdz+xydxdy where S is the bottom side of the triangle with vertices (a,0,0), (0,a,0) and (0,0,a).

This is equivalent to the problem: Find the flux of the vector field \vec F(x,y,z)=\,\lt yz,xz,xy \gt\, through the bottom side of the triangle with vertices (2,0,0), (0,2,0) and (0,0,2), a=2.

We ran the SOLVER first (below) and then used that information to get the graphs here. (Here graph 1 is nothing special since we just have a triangle.)
a=2 
       

Graph 1: We usually will use the parameterization of S that we get below to graph the surface. Here we just graphed the triangle. 
T=polygon([(a,0,0),(0,a,0),(0,0,a)], color='purple', opacity=.7) var ('t') Ax=parametric_plot(vector((t,0,0)),(t,-1,3), thickness=5) Ay=parametric_plot(vector((0,t,0)),(t,-1,3), thickness=5) Az=parametric_plot(vector((0,0,t)),(t,-1,3), thickness=5) show(T+Ax+Ay+Az) 
       

Graph 2: Click and drag to rotate. Notice that the arrows are not all the same length.

We scale the vectors by vs. Notice that the vectors are coming out of the top of the triangle. But our orientation is the bottom of the triangle. The flux is negative!

As always, to get the arrows, we use the parameterization of S to get the "base point" and then use "base point" plus (\vec n \cdot Integrand) to get the "end point".
vs=2 vf=sum([sum([arrow3d((c,d,a-c-d),(c-((c+d-a)*c+(c+d-a)*d-c*d)/vs,d-((c+d-a)*c+(c+d-a)*d-c*d)/vs,a-c-d-((c+d-a)*c+(c+d-a)*d-c*d)/vs), color=hue((c+8)/8), width=0.8) for d in [0..(-c+a),step=0.5]]) for c in [0..a,step=0.5]]) show(T+vf) 
       

Here are the details. 

We parameterize the surface with 2 parameters and find a unit vector \vec n normal pointing down from the triangle.

Work particular to this problem: 

== The equation of the plane going through the 3 points is: x+y+z=a or z=a-x-y. Remember to parameterize an explicit function z=f(x,y), we set x=u, y=v and z=f(u,v)

== We have set a=2

== So: \left\{ \begin{array}{l}x = u \\y = v\\z = a-x-y\end{array} \right.\,\,\,\,\,u \in [0,1 ] \,\,\,\,v \in [0,-u+a]
(To get the intervals, we look at the vertices of the triangle in the u0v plane. They are V1==(u=a,v=0), V2=(u=0,v=a) and V3=(u=0,v=0). We draw this region. It is a triangle where v goes from 0 to the line joining V1 and V2, namely v=-u+a and then u goes from 0 to a.)

== By examination, we see: \vec n=-<1 ,1,1>=<-1,-1,-1>  
(The plane on which the triangle lies is: x+y+z=a. A vector normal to the plane has components equal to the coefficients of (x,y,z) in the plane. So <1,1,1> is a normal vector. We check its orientation. It is pointing up from the top. So we need to multiply by -1 to get the orientation given in the problem.)

We examine the elements of the integral. Remember that the \pm depends on whether or not the parametrization is orientation preserving.

Surface element is the vector product:  \vec S= \vec n \cdot dS = \pm \,\, d \vec S_u \times d \vec S_v \,\,\, du \, dv

Integrand is the mixed product \vec F \,d \vec S = \pm \,\, \vec F \cdot d \vec S_u \times d \vec S_v \,\,\, du\, dv

We define \vec S. We check whether or not our parametrization is orientation preserving. We find the partials of S. We define F and substitute the parameterization.

SOLVER: Our program follows the hand solution method. Everything in red requires your brain!

  1. Parameterize S and input as vector function. Done above.
  2. Find \vec n and input as vector function. Done above.
  3. Input F.
  4. Find the partials of S. 
  5. Determine \pm by checking orientation preserving (+1 if orientation-preserving, else -1). Requires checking that you get \pm 1.
  6. Substitute parameterization in F.
  7. Find the mixed product  \vec F \,d \vec S = \pm \,\,\,\vec F \cdot d \vec S_u \times d \vec S_v \,\,\, du\, dv
  8. Find intervals of integration and integrate. Done above.

Step 0: The program defines our variables. We are given variables (x,y,z). We need (u,v) as parameters.
var ('u v'); var('x y z') 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(u, v\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(u, v\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(x, y, z\right)

Step 1: We define \vec S, \vec n and \vec F. Changes with problem; you must input your parametrization in Sf and your vector n.
Sf=vector((u,v,a-u-v)) n=vector((-1,-1,-1)) F=vector((y*z,x*z,x*y)) 
       

Step 2: The program finds the partial derivatives.
Sprime_u=diff(Sf,u) view(Sprime_u) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(1,\,0,\,-1\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(1,\,0,\,-1\right)
Sprime_v=diff(Sf,v) view(Sprime_v) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(0,\,1,\,-1\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(0,\,1,\,-1\right)

Step 3: The program checks whether our parameterization is "orientation-preserving" or not. We should get 1 or -1. If we get 0 here, we need to change the point, e.g. (u=2.0,v=2.0). Use "real" values with decimal points. (If we get something other than +1, -1 or 0, we have made an error in step 1.)
npar=Sprime_u.cross_product(Sprime_v) np=sign((n.dot_product(npar))(u=1.0,v=1.0)) view(np) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}-1
\newcommand{\Bold}[1]{\mathbf{#1}}-1

Step 4: The program defines a standard sage function for "change of variables" for a 2-variable parametrization (surface).

The program changes the variables of \vec F.
def changevar(f, eqn, newvar1,newvar2): return f.substitute(eqn) 
       
F=changevar(F,x==Sf[0],u,v) F=changevar(F,y==Sf[1],u,v) F=changevar(F,z==Sf[2],u,v) view(F) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left(-{\left(u + v - 2\right)} v,\,-{\left(u + v - 2\right)} u,\,u v\right)
\newcommand{\Bold}[1]{\mathbf{#1}}\left(-{\left(u + v - 2\right)} v,\,-{\left(u + v - 2\right)} u,\,u v\right)

Step 5: Then program finds the mixed product and multiplies it by the orientation from step 3.
M=matrix(([F[0],F[1],F[2]],[Sprime_u[0],Sprime_u[1],Sprime_u[2]],[Sprime_v[0],Sprime_v[1],Sprime_v[2]])) Int=np*M.determinant() view(Int) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}{\left(u + v - 2\right)} u + {\left(u + v - 2\right)} v - u v
\newcommand{\Bold}[1]{\mathbf{#1}}{\left(u + v - 2\right)} u + {\left(u + v - 2\right)} v - u v

Step 6: Then program computes the integral (flux). Changes with problem - we must put in our intervals of integration.
integral(integral(Int,(v,0,-u+a)),(u,0,a)) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}-2
\newcommand{\Bold}[1]{\mathbf{#1}}-2

So our flux is: \iint_S \vec F \,d \vec S = -2  (For variable constant a, we have Flux = \frac{-a^4}{8}).