Answer
$$128\sqrt {17} \pi \ unit^2$$
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 $$S=\int_2^6 2\pi (4x) \sqrt {1 + 16} dx \\=\int_2^6 2\pi (4x) \sqrt {17} dx\\=\int_2^6 8\sqrt {1 7} \pi x dx\\=4 \sqrt {17} \pi [x^2]_2^6 \\= 128\sqrt {17} \pi \ unit^2$$
