Answer
$15 \pi \ unit^2$
Work Step by Step
The area of the surface generated 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:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
Then $$S=\int_0^4 2\pi (\dfrac{3x}{4}) \sqrt {1 + \dfrac{9}{16}} dx \\=\int_0^4 \pi (\dfrac{3x}{8}) \sqrt {16 +9} dx\\=\dfrac{15\pi}{6}|x^2|_0^4\\=15 \pi \ unit^2$$