Surface Integral Homework: A.n dS in Plane 2x+y=6, z=4

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SUMMARY

The discussion focuses on calculating the surface integral of the vector field A = (y, 2x, -z) over the surface defined by the plane 2x + y = 6, constrained within the first octant and capped by the plane z = 4. The user seeks clarification on determining the normal vector and the appropriate projection for the integral. It is established that the normal vector can be derived from the gradient of the plane equation, yielding (2, 1, 0). The integral can be expressed as a double integral of the dot product A · (2, 1, 0) over the area dA.

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Homework Statement



So trying to find the Integral of A.n dS where A is (y,2x,-z) and S is the surface of the plane 2x+y = 6 in the first octant cut off by the plane z=4

Homework Equations





The Attempt at a Solution



So i always solve these by projection...but I am a bit confused this time..

normally the surface is in the form z=f(x,y) so i do z-f(x,y) and take grad to find the normal..

so is the normal vector here just (2,1)? ie. grad 2x-y-6 = 0?

In which case is the integral just the double integral of (y,2x,-4).(2,1,0) dA?

Im a bit confused..Any help would be great!
 
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Any ideas..?
I'm pretty confused about which plane you project onto to solve this...

Thanks!
 
So do you project onto y-z plane?

and is the integral therefore the 1/2 times the double integral of A.(2,1,0) dydz?
 

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