Answer
$\pi \sqrt {c^2+1}$
Work Step by Step
The surface area is given by:
$S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA$
This implies that
$S=\int_{0}^{2 \pi} \int_{0}^{1}\sqrt {c^2+1} r dr d\theta$
Thus, $S=\int_{0}^{2} (\dfrac{1}{2}) \sqrt {c^2+1} d\theta=\pi \sqrt {c^2+1}$
