Answer
$$\dfrac{48143 \pi}{48}$$
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:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
We have: $f(y)=x=4y^{3/2} -\dfrac{\sqrt {y}}{12} \implies f'(y)=6y^{1/2} -\dfrac{y^{-1/2}}{24}$
Use formula: $\int x^n dx=\dfrac{x^{n+1}}{n+1}$
Then $$S=\int_{1}^{4} 2\pi (4y^{3/2} -\dfrac{\sqrt {y}}{12} ) \sqrt {1 + (6y^{1/2} -\dfrac{y^{-1/2}}{24})^2} dy \\=2 \pi \int_{1}^{4}(24y^2+-\dfrac{y}{6}-\dfrac{y}{2}-\dfrac{1}{(12)(24)} \\=\dfrac{48143 \pi}{48}$$
