Answer
$$15 \sqrt {17} \pi$$
Work Step by Step
Let us consider that $f(x)$ is a non-negative function which defines a continuous first derivative on the interval $[a,b]$. The area of the surface obtained when the graph of a function $f(x)$ on the interval $\left[ {a,b} \right]$ is revolved about the $x$- axis can be expressed as:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
Rewrite the function as: $y=4x-1 \implies f(y)=\dfrac{y}{4}+\dfrac{1}{4}$
We have: $f(y)=\dfrac{y}{4}+\dfrac{1}{4} \implies f'(y)= \dfrac{1}{4}$
Then $$S=\int_{3}^{15} \dfrac{\pi}{2} (y+1) \sqrt {1+(\dfrac{1}{4})^2} \ dy \\= \dfrac{\pi}{2} \times \int_3^{15} (y+1) \dfrac{\sqrt {17}}{4} \ dy \\=\dfrac{\sqrt {17} \pi}{8} \times (\dfrac{(y+1)^2}{2}) \\=15 \sqrt {17} \pi$$
