Find the area of $\ln(x^2)$ bound by the $x$-axis, $x=2$, and $x=6$, revolved around the $y$-axis So, I basically know how to do this problem: you use the Method of Washers:
$$V = π(r_o - r_i)^2\delta h $$
to find the area, and enter in the different parts of the equation as I find them.
Here's what I've done:
$$V = π(x-2)^2dy$$
$$V = π\int_2^6(x-2)^2dy$$
$y=ln(x^2), \;\mathtt{so}\; x^2=e^y \;\mathtt{and}\; x=±\sqrt e^y$
$$V = π\int_2^6 (±\sqrt e^y-2)^2 dy$$
Now, the big problem I have with this (unless something is terribly wrong with my work) is that I don't know how to take the definite integral of any plus or minus square root.  Could someone help?
Edit: So, after thinking about it analytically, I realized that the problem probably couldn't be solved by Washers, as the inner and outer arcs wouldn't be touching the same equations constantly.  So, I instead decided to set up my own formula:
$$V_o=V_w-V_i$$ where V_o is the volume of the outer solid, V_i is the volume of the inner hole, and V_w is the theoretical volume of the whole thing if there was no hole.
Putting in the proper equations, we get:
$$V_o=\pi \int_2^6 x^2 \delta y-\pi(2^2)(ln(2^2))$$
which resolves to:
$$V_o=\pi \int_2^6 (±\sqrt e^y)^2 \delta y- ~17.42y$$
and of course becomes:
$$V_o=\pi \int_2^6 e^y \delta y- ~17.42y$$
Finally we get:
$$V_o=1226.7748408036$$
 A: There are two basic approaches one can use: (i) Washers and (ii) Cylindrical Shells. You used washers, but got some details wrong. We use washers, but add a fairly detailed Remark about Cylindrical Shells.
A picture, at least for me, is essential. I will assume that you have a reasonably carefully drawn picture. 
Let us look at a slice of our solid, perpendicular to the $y$-axis. You will see that cross-sections at height $y$ have two different shapes.  
Up to where the curve $y=\ln(x^2)$ meets the line $x=2$, the cross-section is a circle with a cicular hole in it. So the solid is a cylinder with a cylindrical hole in it. We can find the volume of this part by integration, or by basic geometry. The outer cylinder has radius $6$ and height $\ln(2^2)$, the inner cylinder has radius $2$ and height $\ln(2^2)$. So the volume is
$$\pi \ln(2^2)(6^2-2^2).\tag{1}$$
If you really want to find the volume of this part by integration, it will be
$$\int_{y=0}^{\ln(2^2)}\pi(6^2-2^2)\,dy.$$
Now we look at the part of the solid from $y=\ln(2^2)$ to $y=\ln(6^2)$. The washer at height $y$ has outer radius $6$, but now the inner radius is variable, it is $x$, where $y=\ln(x^2)$ and therefore $x=e^{y/2}$. The area of cross-section then $\pi(6^2-(e^{y/2})^2)$. Thus the volume of this part is 
$$\int_{y=\ln(2^2)}^{\ln(6^2)}\pi(6^2-(e^{y/2})^2)\,dy.\tag{2}$$
The integration is easy, we are integrating $\pi(36-e^y)$. 
Now add the two volumes (1) and (2). 
Remark: If we use cylindrical shells, draw a vertical line at $x$, up to our curve. Now draw a vertical line at $x+dx$, where $dx$ is tiny. Rotate the part between $x$ and $x+dx$ about the $y$-axis. We get a cylindrical shell of radius $x$, and height $\ln(x^2)=$. Multiply by $dx$ to approximate the volume of the shell, and "add up" (integrate) from $x=2$ to $x=6$. The volume is
$$\int_2^6 2\pi x \ln(x^2)\,dx.$$
For the integration, rewrite the integrand as $4\pi x\ln x$ and use integration by parts. 
