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 January 9th, 2017, 02:51 AM #1 Newbie   Joined: Dec 2016 From: Tr Posts: 2 Thanks: 0 Vector surface integral for triangle I have question about vector surface integral and I am really stuck here could you please help me? 1- the triangle with vertices (1,0,0), (0,2,0) and (0,0,3) What is the amount of fluid passing through the triangle when the flow vector as given as F=(x,x,x) 2- How can I find this flow vector considering triangular box with vertices (0,0,0), (1,0,0), (0,2,0) and (0,0,3). I mean how can I evaluate flux for each faces of the triangle box for new vertices (0,0,0), (1,0,0), (0,2,0) and (0,0,3).
 January 9th, 2017, 11:09 AM #2 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,158 Thanks: 90 Flow across surface = $\displaystyle \int _{S}$ F.n dS dA=dxdy = k.dS = k.n dS $\displaystyle \int _{S}$F.n dS = $\displaystyle \int _{A}$ $\displaystyle \frac{F.n}{k.n}$dA A is projected area on x,y plane n is surface normal which is conatant
January 10th, 2017, 05:38 AM   #3
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 Originally Posted by kejinatsu I have question about vector surface integral and I am really stuck here could you please help me? 1- the triangle with vertices (1,0,0), (0,2,0) and (0,0,3) What is the amount of fluid passing through the triangle when the flow vector as given as F=(x,x,x)
The three point (1, 0, 0), (0, 2, 0), and (0, 0, 3) lie in the plane given by x+ y/2+ z/3= 1. That should be obvious- each of the points satisfies that equation. That can be written x= 1-y/2- z/3 so we can write parametric equations, in terms of two parameters, s and t, as x= 1- s/2- t/3, y= s, z= t. One vector lying in that plane is the derivative with respect to s, <-1/2, 1, 0>, another is the derivative with respect to t, <-1/3, 0, 1>. The normal vector is the cross product of those two vectors, <1, 1/2, 1/3>. The "vector differential of surface area" is $<1, 1/2, 1/3>dsdt$. The vector <x, x, x>, in these parameters is [tex]<1- s/2- t/3, 1- s/2- t/3, 1- s/2- t/3>[tex] and the integrand is the dot product of those two vectors: $(1- s/2- t/3+ 1/2- s/4- t/6+ 1/3- s/6- t/9)dsdt= (11/6- (11/12)s- (11/1t) dsdt" />

Quote:
 2- How can I find this flow vector considering triangular box with vertices (0,0,0), (1,0,0), (0,2,0) and (0,0,3). I mean how can I evaluate flux for each faces of the triangle box for new vertices (0,0,0), (1,0,0), (0,2,0) and (0,0,3).
The flux is the flow through a surface, here the triangle originally given, with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 3). That has nothing whatever to do with (0, 0, 0) nor with any other faces.

January 11th, 2017, 08:43 AM   #4
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 Originally Posted by zylo Flow across surface = $\displaystyle \int _{S}$ F.n dS dA=dxdy = k.dS = k.n dS $\displaystyle \int _{S}$F.n dS = $\displaystyle \int _{A}$ $\displaystyle \frac{F.n}{k.n}$dA A is projected area on x,y plane n is surface normal which is conatant
1)
Equation of the triangular plane is given by
n.p=d
where n is unit normal to plane and d is distance from origin to plane and
n=aXb/|aXb|
where a and b are any two vectors in the plane, such as (0,2,-1) and (0,3,-1), and d=n.p where p is any vector from the origin to a point on the plane, such as (0,0,1).

To do the flow integral you need z as a function of x,y on the surface to get F(x,y,z)=F(x,y,z(x,y)) on the surface: Solve n.p=d for z=z(x,y).

2)
Repeat 1) to get flow across the other surfaces of the triangular box. On the x,y plane, for example, n=k, and z=0.

F(x,y,z) = f$\displaystyle _{x}$(x,y,z)i + f$\displaystyle _{y}$(x,y,z)$\displaystyle j$+ f$\displaystyle _{z}$(x,y,z)$\displaystyle k$

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### How to find surface integral of a tringle given three vertices

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