Find the surface area of the surface of revolution generated by revolving the graph , around the x-axis.
This is what I got so far I don't know what to do next.
I'm not sure how you got your result ...
If y = f(x) then the surface area produced by revolution of the curve around the x-axis is:
$\displaystyle A = 2\pi \int_a^b\left(y \sqrt{1+(y')^2}\right)dx$
With your example you get:
$\displaystyle A = 2\pi \int_0^8\left(x^3 \sqrt{1+9x^4}\right)dx$
Use integration by substitution:
$\displaystyle u = 1+9x^4~\implies~\dfrac{du}{dx}=36x^3\implies~du=36 x^3 \cdot dx$
Your integral becomes:
$\displaystyle A = 2\pi \cdot \dfrac1{36} \int_0^8\left(36x^3 \sqrt{1+9x^4}\right)dx = \dfrac \pi{18} \int u^{\frac12} du$
Can you take it from here?