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So I am presented with the following problem:

Find the surface area of the cone $z=\sqrt{ x^2 + y^2} $ that lines inside the cylinder $x^2 + y^2 = 2x$.

Im pretty sure a double integral is involved, but I have no clue how to even go about starting this question... any ideas?

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Your first surface isn't a cone... –  J. M. Aug 15 '12 at 12:21
    
@J.M., it seems to be a rather extended habit to call that thing a cone because it looks ridiculously close to a cone, at least for values of $\,x,y\,$ close to zero –  DonAntonio Aug 15 '12 at 15:27
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You can rewrite the formula of the cylinder as $ (x-1)^2+y^2=1 $ so, we have the surface $$S:z=\sqrt {x^2+y^2}, (x,y)\in D$$ wherein $D:=(x-1)^2+y^2≤1$. You know that $$Area(S)=\iint_D d\sigma$$ wherein $d\sigma=\frac{||\nabla f ||}{|\frac{\partial f}{\partial z}|}dx dy$. Here $f=x^2+y^2-z^2=0$ as you noted so, $$d\sigma=\frac{||\nabla f ||}{|\frac{\partial f}{\partial z}|}dx dy=\frac{||(2x,2y,-2z)||}{|-2z|}dx dy$$$$=\frac{\sqrt{4x^2+4y^2+4(x^2+y^2)}}{2|z|}=\frac{\sqrt{8}}{2}dxdxy=\sqrt{2}dxdy$$ Now, I think the rest is easy since your problem is really a homework. :)

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