Creating an integral for finding the volume of this revolution I need to find the volume of a solid that is created by rotating the area within the following boundaries:
$y=x^3$
$y=8$
$x=0$
which is rotated over $x = 3$.
I thought I had the correct integral with 
$$
\int_{0}^{3} (x-3)(8-x^3)dx
$$
but I do not believe that this is correct. How would I correctly get the integral to solve this?

Well, did you draw a picture? You really should for volumes of revolution problems. I will give a few steps here. But your integral is not right. 
Graph the region (bounded by green - $x=3$, red - $y=x^3$, and blue - $y=8$ below), then reflect the region across the axis of revolution, the black graph:

Then you will have a cylinder for which you need to find the volume:
$\pi r^2h$
You're going to do an infinite sum of the volumes, so the integral arrives $\int_{a}^b \pi r^2h$. 
Now you just need to figure out r (hint: green minus red), h (hint: which way are you stacking and then shrinking the cylinders? that's your "$dx$ or $dy$"), and limits of integration (figure out h first!)
-- The integral you wrote is not correct. Why? Compare it to what I'm saying should be your base approach: $\int_{a}^b \pi r^2h$. 
-- I'm also unsure why you were voted down, as this is a reasonable question that you've clearly attempted.
