Volume of a solid I am trying to find the volume of $ y= x^3$ $y=0$ $x=1$ about $x=2$
I know what the graph looks like, I did that part properly. I am just trying to figure out how to calculate the rest of it. I know that I can find the volume of the $x^3$ part but I do not know who to subtract the other part.
 A: Slice your region into thin horizontal slices of thickness $\Delta y$. Each slice, when you revolve it around the line $x = 2$, will become a "washer," that is, a thickened disk with a smaller thickened disk removed (see your picture for a typical slice). The thickness of the disks will be $\Delta y$; what will the areas be?
Well, we need to write them as a function of $y$, so first let's write the cubic as $x = y^{1/3}$. Now, the radius of the big disk will be approximately (EDIT: fixed- see below) $2 - y^{1/3}$ if your horizontal slice includes the point $y$. The radius of the little disk is always 1. So the volume is approximately $\sum \pi((2 - y^{1/3})^2 - 1^2)\Delta y$. Taking a limit as the slices get thinner and thinner, you want the integral
$$
\int_0^1 \pi((2-y^{1/3})^2 - 1)dy
$$
A: We can use the cylindrical shell method, or the slicing (area of cross-section) method. In this case, shells are easier.
$1$. Take a thin vertical strip, of width "$dx$" with base running from $x$ to $x+dx$. This strip is at distance $2-x$, roughly, from the line $x=2$. When you rotate the strip, you get a "shell" of thickness $dx$, height $x^3$, and perimeter $2\pi(2-x)$. "Add up," $x=0$ to $x=1$. We get volume 
$$\int_{x=0}^1 2\pi(2-x)x^3\,dx.$$
$2$. Or use slicing, parallel to the $x$-axis, so we will be integrating with respect to $y$. Take a slice of thickness $dy$, at height $y$. The volume of the slice is $dy$ times the area of cross-section. Note the hole. The area of cross-section is $\pi(2-x)^2-\pi$. In terms of $y$ it is $\pi(2-y^{1/3})^2-\pi$. So the volume is
$$\int_{y=0}^1 \pi\left((2-y^{1/3})^2-1\right)\,dy.$$
Or else you can forget about the hole for a while, integrate $\pi(2-y^{1/3})^2$, and remove the volume of the hole later. The hole is a cylinder, with volume easy to find without calculus.
