Express the area of the given surface as an interated double integral in polar coordinates, and then find the surface area.

A) the portion of the cone z=sqrt(x^{2}+y^{2}) That lies inside the cylinder x^{2}+y^{2}=2x

First thing I did was transform x^{2}+y^{2}=2x into polar coords and got

r^{2}=2rcos(theta)

r=2cos(theta)

So I pretty much guessed that the integral goes from 0≤r≤2cos

and also guessed that theta went from 0≤theta≤pi/2

Not really sure how to prove that is goes from theta to pi/2 just assumed it was in quardrant 1.

Now here is where I get confused,

I then got the partials of z^{2}=x^{2}+y^{2 }doing so I got

dz/dx=rcostheta

dz/dy=rsintheta

I plug it into the formula for surface area which is sqrt[(dz/dx)^{2}+ (dz/dy)^{2}+1]da and im pretty sure this is incorrect.

Thanks for the help