Find area of surface given by
$$ z=x^2-2x+y^2 $$
where
$$ (x-1)^2+y^2 \leq 4 $$
Edit: Initial thought was to use surface-integral, and using the formula
$$ dS = \sqrt{1+(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2}dxdy $$
Doing this, I got:
$$ dS=\sqrt{4x^2-8x+4y^2+5}dxdy $$
but I'm unsure what to do next. Any help with my problem and/or how to solve would be appreciated, thanks.
$$ dS = \sqrt{1+(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2}dxdy $$
Using
$$ \frac{\partial z}{\partial x}=2x-2, \frac{\partial z}{\partial y}=2y, $$
$$ dS=\sqrt{4x^2-8x+4y^2+5}dxdy $$
Change variables to polar coordinate system $$(x,y)=(1+rcosθ,rsinθ)$$
$$ dS=\sqrt{4r^2+1}drdθ $$
Final integral is:
$$ \int_{0}^{2\pi}\int_{0}^{2} \sqrt{4r^2+1}\cdot rdrdθ $$
which gives the final answer:
$$ \frac{\pi}{6}\cdot(17\sqrt{17}-1) $$
Thanks to everyone for their help.