Use the shell method to fine the volume of the solid generated by revolving the region bounded by: $y=12x-11, y=\sqrt{x}$, and $x=0$. 
I know I will be using $V=2\pi\int (\sqrt{x})^2-(12x-11)^2 dx$
It's the setting up of the problem I am having difficulty with. If anyone can help me without giving away the answer, I would greatly appreciate it.
 A: It looks, from your title, as though you need to use the cylindrical shell method for finding the volume of revolution about (I'm assuming here) the $y$-axis. The work you show is more consistent with the disk method (except you'd use $\pi$ in that case).
With the shell method, since volume will be of the cylinder obtained when revolving the region, we need to use as factors: 


*

*$2\pi$, since we revolve the region $360^\circ = 2\pi$ radians (all the way around the y-axis; 

*the radius of the "cylinder" $r(x)$ which will be given simply by "$x$" (the distance of $x$ from the y-axis), and 

*the height of the "cylinder" $h(x)$, which in this case will be the
distance given between the "top curve" minus the "bottom curve". So height will be given by $$h(x) = [\sqrt x - (12x - 11)] = \sqrt x - 12x + 11.$$


One of our bounds of integration will be $x = 0$. The other will be when $x$ is equal to the x-coordinate at which the two curves $y = \sqrt x$ and $y = 12x - 11$ intersect. This will be at $x = 1$. (Recall, we can solve for the $x$ of intersection(s) by equating the two curves: $\sqrt x = 12x - 11$. We can square both sides to get a quadratic equation, solve for the roots, and throw out the negative value. Please confirm that the curves intersect at $(1, 1)$.)
This gives us the following integral to evaluate: $$\int_0^1 2\pi\, r(x) \,h(x)\,dx = 2\pi\int_0^1 x(\sqrt x - 12x + 11)\,dx$$
A: I am assuming you are revolving the region around the y-axis.
If that is the case, you want to find $\int_0^1 2\pi r(x)h(x) dx$, where
r(x) is the distance from a typical vertical line segment to the y-axis and h(x) is the height of a typical vertical line segment. 
