Volumes of unusual solids of revolution in calculus I teach calculus in high school and I was asked two questions today that I wasn't sure how to answer. We were covering volumes of solids and we did a problem involving rotating the region bound by $y=x$ and $y=x^2$ around the y axis and around the line $x=-1$. One of my more curious students asked "What if we rotate around $x=.5$? What if we rotate around the line $y=x$?"
So we started thinking about it and decided that rotating around $x=.5$ is manageable by splitting it up into parts and integrating some with respect to x and some with respect to y using a the washer method for some and the disk method for others. Does that work?
The second question we came up with two ideas for. The first idea is based on the fact that the reflection of $y=x^2$ over $y=x$ is $y=\sqrt{x}$ so we thought about making vertical cross sectional cuts integrating $\pi((\sqrt{x})^2-(x^2)^2)$ but I think the cross sections would not be circular, so vertical cuts wouldn't work. Is this thinking correct?
The second idea was diagonal cuts perpendicular to the line $y=x$ which I think would be circular cross sections, but finding the radius of each would be more complicated because it would require finding the distance along the perpendicular line to the curve. Would this be feasible using this idea? Is this possibly easier using a double integral somehow?
 A: It always surprises me that there are so few figures accompanying questions of solids of revolution. Below is a figure of your area to be rotated, along with its mirror about $x=1/2$. The equation for the blue lines are, of course, $y=x$ and $y=x^2$. Those for the red lines are simply $y=1-x$ and $y=(1-x)^2$. You can challenge your students to find the value of $x$ at the point where the red parabola crosses the blue line. The volume can be computed as the sum of the volumes of the two shaded areas. This completely avoids counting any volume twice. The problem is more complicated of course, but it will show your students that there's nothing mystical about it. Alternatively, one can construct the single area shown in second image below, but that would be more difficult to compute with the standard methods, but it makes the same point.
For your information, I would calculate the centroid of the second figure (relative t0 $x=1/2$) and use Pappus's centroid theorem. I always use Pappus's theorem as I think these disk, washer, shell. and whatever methods confine the thinking.


A: To write the equation of the parabola in the axes $x'=(y+x)/\sqrt{2}$,$y'=(y-x)/\sqrt{2}$, should not be difficult, of course if (elementary) change of axis is part of the course.
The rotation around $x=1/2$ is by $\pi$ radians, supposedly. Then the method you suggest (part in $x$, part in $y$) is viable. But why not calculate both parts as tiny washers in $dx$, with the height given by $y=x$ taken as positive and that of the parabola taken as negative ?   
However the most interesting aspect, I think, would be to compare the volumes obtained around $x=0$, $x=-1$, etc. by the integration we are talking about, and then as cross-area $\times$ centroid path length. 
