Given the parameterization
[tex]\vec r(u,v) = u\cos(v) \, \vec\imath + u\sin(v) \,\vec\jmath + v \,\vec k[/tex]
take the normal vector to be
[tex]\vec n = \dfrac{\partial\vec r}{\partial u} \times \dfrac{\partial \vec r}{\partial v} = \sin(v) \,\vec\imath - \cos(v) \,\vec\jmath - u \,\vec k[/tex]
(The order of partial derivatives in the cross product doesn't matter since this a scalar line integral.)
Compute the magnitude of the normal vector.
[tex]\|\vec n\| = \sqrt{\sin^2(v) + (-\cos(v))^2 + (-u)^2} = \sqrt{1+u^2}[/tex]
so that the area element reduces to
[tex]dS = \|\vec n\| \, du\,du = \sqrt{1+u^2}\,du\,du[/tex]
Evaluate the integrand at [tex]\vec r[/tex] to get
[tex]\sqrt{1 + (u\cos(v))^2 + (u\sin(v))^2} = \sqrt{1 + u^2}[/tex]
The surface integral reduces to
[tex]\displaystyle \iint_S \sqrt{1+x^2+y^2} \, dS = \int_0^{2\pi} \int_0^1 (1+u^2) \, du \, dv = 2\pi \int_0^1 (1+u^2) \, du = \boxed{\frac{8\pi}3}[/tex]
