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January 9th, 2017, 03:51 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, 12:09 PM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91 
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, 06:38 AM  #3  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,876 Thanks: 766  Quote:
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January 11th, 2017, 09:43 AM  #4  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,194 Thanks: 91  Quote:
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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