Answer
$$96 \pi$$
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: $f(y)=x=\sqrt {12y-y^2} \implies f'(y)=\dfrac{6-y}{ \sqrt {12y-y^2}}$
Then $$S=\int_{2}^{10} 2\pi (\sqrt {12y-y^2}) \sqrt {1 + (\dfrac{6-y}{ \sqrt {12y-y^2}})^2} dy \\=2 \pi \int_{2}^{10} \sqrt {12y-y^2+(6-y)^2} \ dy \\=2\pi \int_{2}^{10} (6) \ dy \\=12 \pi[y]_2^{10} \\=96 \pi$$
