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
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
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
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
thanks PF :D