Due to the symmetry of the paraboloid about the z-axis, you can treat this is a surface of revolution. Consider the curve [tex]y=x^2[/tex], with [tex]1\le x\le2[/tex], and revolve it about the y-axis. The area of the resulting surface is then
[tex]\displaystyle2\pi\int_1^2x\sqrt{1+(y')^2}\,\mathrm dx=2\pi\int_1^2x\sqrt{1+4x^2}\,\mathrm dx=\frac{(17^{3/2}-5^{3/2})\pi}6[/tex]
But perhaps you'd like the surface integral treatment. Parameterize the surface by
[tex]\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k[/tex]
with [tex]1\le u\le2[/tex] and [tex]0\le v\le2\pi[/tex], where the third component follows from
[tex]z=x^2+y^2=(u\cos v)^2+(u\sin v)^2=u^2[/tex]
Take the normal vector to the surface to be
[tex]\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath-2u^2\sin v\,\vec\jmath+u\,\vec k[/tex]
The precise order of the partial derivatives doesn't matter, because we're ultimately interested in the magnitude of the cross product:
[tex]\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|=u\sqrt{1+4u^2}[/tex]
Then the area of the surface is
[tex]\displaystyle\int_0^{2\pi}\int_1^2\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\int_0^{2\pi}\int_1^2u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
which reduces to the integral used in the surface-of-revolution setup.
