Answer
$$\pi \sqrt {c^2+1}$$
Work Step by Step
Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$
We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$
$$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA \\=\int_{0}^{2 \pi} \int_{0}^{1}\sqrt {c^2+1} r \space dr \space d\theta\\ =\int_{0}^{2} (\dfrac{1}{2}) \sqrt {c^2+1} \space d\theta \space \\=\pi \sqrt {c^2+1}$$
