Answer
$$\dfrac{49 \pi}{3} $$
Work Step by Step
We integrate in order to find the surface area:
$$S= \int_{c}^{d} (2\pi y) \sqrt {1+(y')^2}$$ $$\\ =\int_{2}^{6} (2 \pi ) \times \sqrt {y} \times \sqrt {\dfrac{4y+1}{4y}}dy \\ =(\dfrac{\pi}{4}) \times [\dfrac{2(4y+1)^{3/2}}{3}]_2^6 \\= \dfrac{49 \pi}{3} $$
