Set up the integral for the volume obtained by rotating the region bounded by the curves y = x, y = 0, x = 4, and x = 6 about x = 1 I've been stumped on this problem for hours because it looks like a weird version of what I am used to.
I have no idea how to set this integral up. Here is what the graph looks like (I know its a y=x graph but still here's the visual: https://i.gyazo.com/b7468a8b197e3e9c0403b7940867f4a3.png)
BUT I only want the volume from 4 to 6... but this problem seems like the one where you use y values instead of x values for the bounds of integration
So is it ∫ a = 4, b = 6?
should my a = 0, and my b = 6? I just don't get how to set up such a strangely shaped problem. 
If I am using the disk method, I would be doing pi∫(outer)^2 - (inner)^2 dx or dy
but what would even be the outer/inner function in this scenario? Like I have no idea. X = 1 is a VERTICAL line so I can't take the "highest" function and make it the outside, and the "lowest" the inner. I'm supposed to (according to what I have been taught) take the farthest and subtract it from the closest. I can't tell how you would know what the farthest function from x = 1 would even be.
I DO know that I need to do (outer - 1)^2 and (inner - 1)^2 in this scenario because we are going about x = 1, not the y axis itself. 
Edit: looks like I was making it way harder trying to do the Disk method when I could've just done Cylindrical Shell Method
 A: With the shell method it's
\begin{equation}
\int_4^6 2\pi rh\,dx
\end{equation}
with $r=x-1, h=x$
so
\begin{equation}
\int_4^62\pi (x^2-x)\,dx
\end{equation}
A: The fact that you say "should my a = 0, and my b = 6" without even having mention what "a" or "b" are indicates to me that memorizing formulas without thinking about where those formulas come from.  You should be thinking "since I am rotating around the x= 4 axis, you can think of each "layer" (for each value of y) as a disk with center at x= 1.
Now, there is a problem here.  From y= 0 to y= 1, we have an entire rectangle, with vertices at (1, 0), (4, 0), (1, 1), and (4, 1) rotates around x= 1.  From y= 0 to y= 4, only the triangle with vertices at (1, 1), (4, 1), and (4, 4) rotates.  It will be simplest to do those two regions separately.  
For the lower, rectangular part, for every y from  0 to 1, the part rotated about x= 1 is a line segment from x= 1 to x= 4 which rotates into a circle with radius 4- 1= 3 which has area $\pi r^2= 9\pi4$.  Thinking of each such circle as a disk with thickness dy (dy NOT "dx" because we are moving [b]up[/b] the vertical x= 1 line so the limits of integration will be from y= 0 to y= 1, NOT x= 0 to 4 or x= 1 to 4.) and so a disk with volume 9\pi dy.  The integral, using "disks" will be $\int_0^1 9\pi dy= 9\pi\int_0^1 dy= 9\pi$.  We could also have done that without integrating by noting that, geometrically, that is a cylinder with radius 9 and height 1.
For the upper part, for every y from 1 to 4, we have a "washer" which we can think of as circle with an inner circle missing.  The outer radius is from x= 1 to x= 4, so 3, as before.  The inner radius, for each y, is from x= 1 to x= y so y-1.  The area of the entire circle is, as before, $9\pi$  while the inner area is $\pi(y- 1)^2$.  The area of the "washer" is $9\pi- \pi(y- 1)^2= \pi(8+ 2y- y^2)$.  The volume of such a "washer" is that area times the thickness, dy.  The volume of the entire part, from y= 1 to y= 4 is $\int_1^4 \pi(8+ 2y- y^2)dy= 18\pi$.  The volume of the whole figure is the sum of these $9\pi+ 18\pi= 27\pi$.
You can do this as one single integral, integrating with respect to x, using the "shell method".  For each value of x, from 1 to 4, we imagine the  vertical line rotated around x= 1 so that it sweeps out a cylindrical shell of radius x- 1 and height y: the area is the circumference of that cylinder times its height, $2\pi(x-1)y= 2\pi(x- 1)x= 2\pi(x^2- x)$.  The volume is that area times the thickness of shell, dx (since each shell shifts along the x-axis): $\int_1^4 2\pi(x^2- x)dx= 27\pi$.
A: I always like transform these problems so the rotation is about the x-axis (y=0). So two transformation must occur. First invert everything, then shift down 1.
original: 
rotation about: x=1 
bounded by: y=x, y=0, x=4,x=6
equations after inverting: 
rotation about: y=1 
bounded by: x=y, x=0, y=4,y=6
then equations shifting down 1: 
rotation about: y=0 
bounded by: x-1=y, x=0, y=3,y=5
This is the exact same shape with the same volume, but reoriented around the x-axis. 
Now we have a new image:
 
Notice we have intersections at (4,3) and at (6,5)
And I would use a 2 part washer method here.
A's volume is 
$A = \pi \int_{0}^{4} 5^2 - 3^2 dx$
B's volume is 
$B = \pi \int_{4}^{6} 5^2 - (x-1)^2 dx$
thus the volume is: 
$Vol= \pi \int_{0}^{4} 5^2 - 3^2 dx +  \pi \int_{4}^{6} 5^2 - (x-1)^2 dx$
If want to back to a vertical rotation , just change the x's to y's.
$Vol= \pi \int_{0}^{4} 5^2 - 3^2 dy +  \pi \int_{4}^{6} 5^2 - (y-1)^2 dy$
