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I'm not sure what you're doing here, but it's true that in cylindrical coordinates (R,Ψ,z), the Euclidean metric takes the form ds² = dR² + R²dΨ² + dz², just as you have it, but the question was about a surface of constant R. Obviously, dR = 0 on that surface, say R = a.

Substitution's all well and good, but perhaps you're missing something important. The metric by definition finds inner products of tangent vectors, and you know from the first part the tangent vectors Ψ (= RΨ^, not normalized) and z corresponding to the surface. Thus, with those two as the basis,
g11 = Ψ·Ψ = R²; g12 = Ψ·z = 0 = g21; g22 = z·z = 1.
Therefore, you can simply read off the metric components from the usual
ds² = g11 dΨ² + g12 dΨdz + g21 dzdΨ + g22 dz²,
since all the relevant partial differentiation was already done in finding the orthonormal basis.


Yes, except for a factor of a², because the absolute value of a vector is just another name for the norm (|x| = sqrt(x·x)). I'm not sure what you're asking if you've already found the cross correctly.


I presume that you mean for a function f:R³→R, the surface is defined by f = c for some constant c, which would make the statement correct. One way to think of this is to continue interpreting the gradient as the proper generalization of derivative in higher dimensions (as we've done in the previous thread regarding integrals), so that what we really have is a chain-rule type of situation. To be more explicit, if you have think of a local patch of the surface as having the form (x,y,z) = F(u,v), then f(F(u,v)) = c differentiated gives ∇f·∂F/∂u = 0, but ∂F/∂u is just the surface tangent vector in the u-direction. Since the parametrization in terms of (u,v) is arbitrary, ∇f must orthogonal to every tangent vector of the surface.

Perhaps it'll be easier to visualize if you consider a curve γ on the surface that has constant coordinates y,z (i.e., only x is allowed to vary: γ = (X(t),Y,Z)). Then f(γ) = c differentiated is just makes use of the ordinary chain rule. Now consider what happens for a general curve where the y and z coordinates are allowed to vary--there are a now partials in those directions as well.


Simple answer: because that's what substitution gives you. Just differentiate: dx = dR cos Ψ - R sin Ψ dΨ, dy = ... and substitute in dx dy.
Deeper answer: because that's the volume element, which is the determinant of the metric tensor, and the determinant of the metric in those coordinates is RdRdΨ. Note: "volume" is a bit of a misnomer in the 2-dimensional case, since it's really area, but I'm using it in the general sense.
Intuitive answer: Start from some point with coordinates (R,Ψ) and plot (R,Ψ+dΨ), (R+dR,Ψ), and (R+dR,Ψ+dΨ). This is a curved region that's rectangular in the infinitesimal limit, with one side dR and another side along the circumference RdΨ. Hence the area element is RdRdΨ in polar coordinates. If you do the same thing in Cartesian coordinates, we would get dxdy. Since areas of regions are independent of which coordinates we use, they better be equal.


The notation you've used there doesn't make much sense to me, so I'm not sure what you're doing here. However, that looks like a paraboloid of revolution: r²+z-1 = 0 in cylindrical coordinates. You're probably going to end up trying to make it into an integral over a region in the xy plane (dA = dxdy = rdrdΨ).

Or, you can probably just tie everything together and look at the induced metric over the surface: ds² = sin²u dv² + [5-4cos²u]cos²u du². Hence the surface area element is just the determinant of the metric, dS = [5-4cos²u]sin²u cos²u du dv, and you can get a surface normal with that cross product you were doing earlier, but this method is a lot messier. You should probably double-check that, though, because I'm too lazy to do so.


It's in the direction of the greatest increase of f, with the magnitude being the slope in that direction. In general, ∇f·r is the ordinary derivative (slope) in the r direction.


I usually try to picture those differentials as pairs of infinitesimal planar regions. I can try to explain the picture later in more detail if you want, but right now I've done too much typing already.