Finding the Normal Vector for a Triangle in the Plane Using Stoke's Theorem

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The discussion focuses on evaluating the line integral ∫C F·dr for the vector field F(x,y,z) = y²i + z - xk over the oriented triangle C in the plane defined by the equation 2x + y + z = 4. The normal vector, denoted as , is calculated using the gradient function g(x,y) = -2x - y + 4, resulting in = <-2, -1, 1>. The final computed answer for the integral is 4/3, although there is confusion regarding the correct components of the normal vector.

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Homework Statement



Let C be the oriented triangle lying in the plane 2x+y+z=4. Evaluate ∫
[itex]_{C}[/itex] F*dr where F(x,y,z)=y2i + z - xk.

Homework Equations



I will be using ∫C∫curl F [itex]\bullet[/itex] [itex]\vec{n}[/itex] to solve this problem. But when I'm trying to find [itex]\vec{n}[/itex] using -gxi - gyj + k, I get

Using g(x,y)= -2x -y + 4
[itex]\vec{n}[/itex] = < -2, -1, 1>

Did I do that correctly? Will the k just be 1 or 4?

My final answer was 4/3?
 
Last edited:
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aaronfue said:

Homework Statement



Let C be the oriented triangle lying in the plane 2x+y+z=4. Evaluate ∫
[itex]_{C}[/itex] F*dr where F(x,y,z)=y2i + z - xk.

Homework Equations



I will be using ∫C∫curl F [itex]\bullet[/itex] [itex]\vec{n}[/itex] to solve this problem. But when I'm trying to find [itex]\vec{n}[/itex] using -gxi - gyj + k, I get

Using g(x,y)= -2x -y + 4
[itex]\vec{n}[/itex] = < -2, -1, 1>

Did I do that correctly? Will the k just be 1 or 4?

My final answer was 4/3?

But gx=(-2) so -gx=2. Why are you writing <-2,-1,1>? Shouldn't it be <2,1,1>?
 

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