# Homework Help: Flux, how to find n (normal) and derivation of formula?

1. Aug 8, 2012

### Gridvvk

1. The problem statement, all variables and given/known data
Find the flux of the following fields:
F_1 = xi + yj
F_2 = -yi + xj

across the following curve: The circle r(t) = (cost) i + (sint) j
t is from [0,2pi]

2. Relevant equations

Flux = ∫F dot n ds = ∫M dy - N dx

3. The attempt at a solution

For F_1 I got:
M = x = cos t
N = y = sin t
dy = cost t dt
dx = -sin t dt

Flux for F_1 = ∫[0,2pi] cos^2 t + sin^2 t dt = ∫[0,2pi] dt => 2pi

For F_2 I got:
M = -y = -sin t
N = x = cos t
dy = cos t dt
dx = - sin t dt

Flux for F_2 = ∫[0, 2pi] -costsint + costsint dt = ∫0 = 0

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My answers are correct (can someone verify if method was used correctly?), but the answer key used a different method:
It said "n = <cost , sin t>, and proceeded to dot that with each field.

My concern was how n was found, I thought n = T' / |T'|, where T = r' / |r'| :
so: r' = <-sint, cos t> = T (because |r'| = 1)
T' => <-cost, -sint> = n (because |T'| = 1 as well), so how did they get their n?

Also, I'm a bit unsure on how Flux = ∫M dy - N dx is derived -- in the textbook they did:
n = T x K = (dx / ds i + dy/ ds j) x k = (dy / ds i - dx/ ds j), but if someone can spell that out for me I'd appreciate it.

Thanks

2. Aug 8, 2012

### szynkasz

Formula $n=\frac{T'}{|T'|}$ is incorrect, try parabola $r(t)=[t,t^2]$. Vector product $T\times k$ is perpendicular to $k$ which means it lies in $XY$ plane. It is also perpendicular to $T$ so it is normal to the curve.

3. Aug 8, 2012

### Gridvvk

Is the normal vector here in the flux case, different from the "unit vector normal" ? If so, how is n for flux calculated then?

4. Aug 10, 2012

### szynkasz

No, it is not different, except $T\times k$ is not unit vector, so $n=\frac{T\times k}{|T\times k|}$

5. Aug 10, 2012

### LCKurtz

They just observed that the curve was a circle and the position vector to the circle is perpendicular to it.

6. Aug 10, 2012

### Gridvvk

Oh that makes sense, thanks for the clarifications szy and LC.