Find the area of the surface.
The portion of the cone z = 2(x^2 +y^2)^1/2 inside the cylinder x^2 + y^2 = 4.
you're looking for the surface area of the part $\displaystyle z=2\sqrt{x^2+y^2}$ and $\displaystyle (x,y)$ was given from the disk $\displaystyle x^2+y^2\le4.$
compute $\displaystyle f_x,f_y$ so the double integral is $\displaystyle \iint\limits_{x^{2}+y^{2}\le 4}{\sqrt{1+\left( f_{x} \right)^{2}+\left( f_{y} \right)^{2}}\,dA},$ and the bounds in polar coordinates are $\displaystyle 0\le r\le2,$ $\displaystyle 0\le\varphi\le2\pi.$