Find the area of the conical shell that has an equation $ z=\sqrt{x^2+y^2} $ using polar coordinates

Question

I need to find the area of the conical shell that has an equation $$ z=\sqrt{x^2+y^2}, $$ where $0 \leq a < b$ and $a \leq z \leq b$

What I have found by far is that our $z=r$, thus $a \leq r \leq b$ and $0 \leq \theta \leq 2\pi$

Which means that I will have to integrate $$ \int_0^{2\pi} \int_a^b r^2drd\theta $$

Please correct me if I made a mistake. I feel like either the bounds of integration or the function specification might be wrong.

Answer 1

$$\text{area(A)}=\iint_A\sqrt{1+z_x^2+z_y^2}dxdy$$ where $$\sqrt{1+z_x^2+z_y^2}=\sqrt{1+\left(\dfrac{2x}{2\sqrt{x^2+y^2}}\right)^2+\left(\dfrac{2y}{2\sqrt{x^2+y^2}}\right)^2}=\sqrt{2}$$ then $$\text{area}=\int_0^{2\pi} \int_a^b r\sqrt{2}drd\theta$$