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## Homework Statement

Calculate ∫∫ f(x,y,z)dS for the surface G(r,θ) = (rcosθ,rsinθ,θ), 0<r<1, 0<θ<2pi.

f(x,y,z) = (x^2+y^2)^(1/2) = r

## Homework Equations

## The Attempt at a Solution

so the surface is given so I have to find the normal vector...

G_r = cos(θ),sinθ,0

G_θ = -rsin(θ),rcos(θ),1

G_r = G_θ = normal vector = sin(θ)I - cos(θ)j + rk

magnitude of normal vector

=

(1+r^2)^(1/2)

heres my main question, usually if I parameterize in terms of u and v this would be like

dS = |Normal vector| du dv

but since this is in terms of r and θ do I just do

dS = |Normal vector| dr dθ

or

dS = |Normal Vector| r dr dθ

cause earlier in this course when we went from rectangular coordinates to polar we had to add an r, does that still apply here?

So basically would my final integral be

∫∫r(1+r^2)drdθ

or

∫∫r^2(1+r^2)drdθ

thanks PF :D