Divergence theorem

[boreal sorrel]

How to get N2
The outer normal vector on Σ2
I think the answer is wrong

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[hexed oak]

Note that the normal vector to a surface f(x, y, z) = C is ∇f. So, in our case:
f = x^2 + y^2 = 4
∇f = 2{x, y}
So, now you can find the unit normal vector.

[cedar cosmos]

I think OP is right actually

I mean we all agree on the direction of this normal vector, but the amplitude bothers me

[hexed oak]

Oh, I see what you mean.

[cedar cosmos]

because it's not... well, normalized

and second of all, even if you leave it like that, the computation of the integral is definitely wrong

F . n = 8 cos(u) * 3 cos(u) - 12 sin(u) * 3 sin(u)

= 24 cos²(u) - 36sin²(u)

integrating wrt v will yield a factor 2

but as a chemist/physicist you're aware that the mean value of a cos² or sin² on one period is 1/2, so the integral wrt u is pi

so integrating that will yield 24 pi - 36 pi = -12 pi

multiplied by 2, you still don't get the result

[hexed oak]

Well, that yields 3 times the result they showed, it seems.

Which corresponds to the factor of 3 they forgot to remove.

[cedar cosmos]

no, even accounting for the factor 3 you're still not good

i used the factor 3 and got -12 pi

so without it's -4 pi

not -54pi, or -18pi

[hexed oak]

Hm...

Let me try.

[cedar cosmos]

sure sure lmk how it goes

maybe i tripped somewhere cause i only did mental calc

[hexed oak]

Nah, getting confused on the lateral surface. Haven't done this in a while.

[cedar cosmos]

it's just the outer surface of a cylinder

the round one

so the normal vector is obviously radial

if (x, y, z) is on it, one radial vector is just (x, y, 0)

but you've gotta normalize it as well

(it coincides with the gradient thing you've said earlier)

[boreal sorrel]

I think the answer of Σ2 part is wrong

[hexed oak]

That's obvious. My trouble, however, is with dS.

[cedar cosmos]

it's du dv

du = small variation in angle

dv = dz

[hexed oak]

Just that? Hm...

What about the jacobian?

[cedar cosmos]

well I can check

[hexed oak]

It isn't a flat surface.

[cedar cosmos]

I might be wrong after all

You're right, I am in fact incorrect

because the radius, although constant, is 2

but it does not suffice to show that the factor of proportionality is not 1

[cedar cosmos]

In the first place, you can't really compute the elementary surface using cartesian coordinates

otherwise you'd struggle with something along the lines:
x = 2 cos(u)
dx = -2sin(u) du
y = 2 sin(u)
dy = 2 cos(u) du
(du)² = (1/2²)(dx)² + (1/2²)(dy)²

Hmm...

I think that dS = 2 du dv actually

Or in general, an elementary arc length will be given by root(dx² + dy²)

[cedar cosmos]

But this shows that du is 1/2 times the elementary length

so 2 du is the elementary arc length

[hexed oak]

Well, you can parametrize a surface. However, I don't remember the full approach.

[cedar cosmos]

For the same reason why you substitute dxdy = r dr dtheta usually

or I guess, dr * (r dtheta)

variation in radius * variation in arc length

Here we're pretty much doing the same thing, but with a constant radius

and instead of dr, we take dz

that's the physicist approach

now to express that mathematically, I have to admit that I'm struggling

[gilded hatch]

yea but i dont think that changes anything 0<= r <= 2, 0<= theta <= 2pi and still results in -8pi

i think this is what it should look like
$$ \int_0^{2 \pi } \int_0^2 \int_0^2 -r dr dz d\theta$$

[wintry kite]

L

[cedar cosmos]

No it's clearly not

We are not computing a volume but a surface

Your infinitesimal element is homogeneous to a volume

[gilded hatch]

yes but if we use divergece theorem surface integral changes to volume of div F

[cedar cosmos]

But that is hardly the point of the question

The question is about the correctness of the lateral surface area in the key

[gilded hatch]

all im saying is that changing to polar coordinates still leads to -8pi and not -54pi i think that was part of the quesiton 😁

[cedar cosmos]

Which confirms my point that the computation in the key is probably wrong

As said by the original poster

[gilded hatch]

yes

[cedar cosmos]

Sorry if I sound a bit cranky because I am beyond pissed right now (not your fault), but I guess we both align by this opinion

That the key is incorrect I mean

[gilded hatch]

no no its okay