Find the surface area of the cone z=4sqrt(x^2+y^2) and above a region in the xy-plane with area 3.

I tried this problem and here's what i've done------

r=sqrt(x^2+y^2)

The slope s is s=sqrt(r^2+16r^2)=sqrt(17)r

A surface=pir^2+pi r s=pi r^2(1+sqrt17)

3=pi r^2(1+sqrt17) ===> r=0.4317

this is as far as i went, but what do I do next to get the surface area?

Thanks in advance!!!