2
$\begingroup$

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?

  • 0
    @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 zero2012-08-15

1 Answers 1

2

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. :)