Answer
$$\dfrac{8\pi}{3} (5\sqrt 5-2 \sqrt 2)$$
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: $y=\dfrac{x^2}{4} \implies f(y)=2y^{1/2}$
We have: $f(y)=2y^{1/2} \implies f'(y)=2 \times \dfrac{1}{2 \sqrt y}$
Then $$S=\int_{1}^{4} 4\pi \sqrt {y+1} \ dy \\=4 \pi [\dfrac{2}{3}(y+1)^{3/2}]_1^4\\=4 \pi \times \dfrac{2}{3} \times [(1+4)^{3/2}-(1+1)^{3/2}] \\=\dfrac{8\pi}{3} (5\sqrt 5-2 \sqrt 2)$$