Answer
$$320 \pi$$
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:
$$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$
Rewrite the function as: $x^2+y^2=100 \implies y=\sqrt {100-x^2}$
We have: $f(x)=y=\sqrt {100-x^2} \implies f'(x)=\dfrac{-x}{2 \sqrt {100-x^2}}$
Then $$S=\int_{0}^{8} 4\pi (\sqrt {100-x^2}) \sqrt {1 + (-\dfrac{x}{\sqrt {100-x^2}})^2} dx \\=4 \pi \int_{0}^{8} \sqrt {100-x^2+x^2} \ dx \\=4\pi \int_{0}^8 (10) \ dx \\=40 \pi[x]_0^8 \\=320 \pi$$