Find volume by rotation $y = x^3, y = 0, x = 1;$ about $x = 2$
I understand that this is a cross sectional graph and I know how to do it with two curves, but the $y = 0$ throws me off. I am left with one curve so I assume the integral would be:
$\int \pi(2-x^3)^2dx$
but I'm not getting the correct answer. Maybe I calculated wrong. Could someone assist me with this?
 A: The region is not really well described, but it it likely that you are to rotate the finite region below $y=x^3$, above $y=0$, and to the left of $x=1$  about the line $x=2$.
If that is the case, the method usually called the method of cylindrical shells is a natural approach. Take a thin vertical slice of width going from $x$ to $x+dx$. this slice has height about $x^3$. When it is roated about $x=2$, we get a shell of radius $2-x$, with volume $2\pi(2-x)x^3\,dx$. "Add up" (integrate) from $x=0$ to $x=1$. We get
$$\int_0^1 2\pi(2-x)x^3\,dx.$$
This is probably $\frac{3\pi}{5}$.
We could alternately use the method of slicing (cross-sections). In that case we will be integrating with respect to $y$. At height $y$, the cross-section is a circle of radius $2-x$, with a hole of radius $1$ drilled out. We have $x=y^{1/3}$, so our volume is 
$$\int_{y=0}^1 \pi\left((2-y^{1/3})^2-1^2\right)\,dy.$$
The algebra is somewhat more unpleasant, but again we get $\frac{3\pi}{5}$.
Remark: The formula proposed in the post is sort of related to rotating about the line $y=2$. For that, we would want $\int_0^1 \pi\left(2^2-(2-x^3)^2\right)\,dx$. 
It is usually not very hard to solve these rotation problems, if we draw a picture and each time go back to basics. Using a remembered formula is, for me at least, far less reliable. 
A: Well, first you need to define the area that is to be rotated.  You do need all three curves to do that.  It's a good idea to sketch them very roughly...
The area is bounded on the bottom by $y=0$, on the right by the vertical line $x=1$. The third "side" is the curve $y=x^3$.  So, the three "corners" of the area are $(0,0)$, $(1,0)$, and $(1,1)$.
Next, what is the axis of rotation?  Easy, it's the vertical line $x=2$.
Now, you need to decide whether to use shells or discs;  let's use discs.  Or rather washers,  The tricky bit is that the axis of rotation is distinct from the area being rotated.  The final volume is basically a volcano cone minus the crater...
The outer radius of the washer is $2-x$, the inner radius is $2-1=1$, the thickness is $dy$, so the volume of the washer is $$pi((2-x)^2-1^2)dy$$
Replace x with $y^{(1/3)}$ and add up all the washers from $y=0$ to $y=1$
General question:  is this too much help?  I'm new here...
