Answer
$$48 \pi$$
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:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
We have: $f(x)=y=\sqrt {8x-x^2} \implies f'(x)=\dfrac{8-2x}{2 \sqrt {8x-x^2}}$
Then $$S=\int_{1}^{7} 2\pi (\sqrt {8x-x^2}) \sqrt {1 + (\dfrac{8-2x}{2 \sqrt {8x-x^2}})^2} dx \\=2 \pi \int_{1}^{7} \sqrt {8x-x^2+(4-x)^2} \ dx \\=2\pi \int_{1}^7 \sqrt {16} \ dx \\=8 \pi[x]_1^7 \\=48 \pi$$
