If you fix a point on the curve in the given interval, and revolve that point about the [tex]x[/tex]-axis, it will trace out a circle with radius given by the function value [tex]y[/tex] for that point [tex]x[/tex]. The perimeter of this circle is then [tex]2\pi(8\sqrt x) = 16\pi \sqrt x[/tex].
The surface in question is essentially what you get by joining infinitely many of these circles at every point in the interval [0, 9].
So, the surface area is given by the definite integral
[tex]\displaystyle \int_0^9 16\pi \sqrt x \, dx = 16\pi\times\frac23 x^{3/2}\bigg|_{x=0}^{x=9} = \frac{32\pi}3 \left(9^{3/2} - 0^{3/2}\right) = \boxed{288\pi}[/tex]
