# Find area of surface

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.

• 1) Do you mean volume or surface area? 2) What attempts have you already made in solving this problem? – Emory Sun May 3 at 15:59
• Welcome to MSE. Do you have any ideas of how to start? Do you know what surface area is, or do you have any tools about how to find it? Do you know anything about surface integrals, parametrized surfaces, or related ideas? Please edit your question to improve it; as it stands, it's an off-topic do-my-problem-without-me post. – user296602 May 3 at 15:59
• I've edited my post with what I've tried so far. I think I've found dS, but I'm not sure what to do next. Thanks. – Martin Pham May 3 at 16:03
• Hint: change variable to a shifted polar coordinate system $(x,y) = (1 + r\cos\theta, r \sin\theta)$. – achille hui May 3 at 16:08

$$dS = \sqrt{1+(\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y})^2}dxdy$$

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:

$$\int_{0}^{2\pi}\int_{0}^{2} \sqrt{4r^2+1}\cdot rdrdθ$$

$$\frac{\pi}{6}\cdot(17\sqrt{17}-1)$$