Solid of revolution about a slanted line I just thought about this idea and I decided to work on it. 
After taking on a general case, which proved to be too difficult, I tried a specific case. Something simple like the curve $y_1 = x^2$ rotating about the line $y = x$
Which is the same as rotating $y = \sqrt{x}$ about the x-axis.
I know I need to find the new radius which is the line perpendicular to y = x and I need to pick a particular point on the curve and the line.
So if i were to pick say, x = 0.5, the perpendicular line would be
$y =-x + 1$
So my solid of revolution integration would be
$\pi \int_{a}^{b} (-x + 1)^2 d?$
Unfortunately it proved to be very difficult to find the slanted differential in terms of dx and I couldn't figure out what the change of variables of bounds were.
Any ideas?
 A: One approach is to rotate the parabola $45^\circ$ clockwise and then do the volume calculation as usual.  The result is mildly messy, but with care it can be made to look nice.
If we don't want to rotate, and want to integrate with respect to $x$, here is an approach that I think works. If some details are wrong, someone will soon point out the correction.  Please note that the formulas below assume that $x$ is positive. 
The distance from the point $(x,x^2)$ on the parabola to the line y=x$ is, by the usual distance to a line formula. equal to 
$$\frac{|x-x^2|}{\sqrt{2}}.$$
Look at little slices as usual, going from $x$ to $x+dx$.  If we can express the width of the slice as $w(x)dx$, then the volume we are looking for is
$$\int_0^1 \frac{\pi}{2}(x-x^2)^2 w(x)\,dx.$$
Now we need to go after $w(x)$.  So we will compute the "slanted differential" mentioned in your post. Let's do this engineering-style.  Project the point $(x,x^2)$ onto the line $y=x$. Let the result be the point $P$.  Also, project the point $(x+dx, (x+dx)^2)$ onto the line $y=x$. Let the resulting point be $Q$. We want the distance between $P$ and $Q$. After a while I got (discarding higher-order infinitesimals as usual) that this distance is
$$\frac{1+2x}{\sqrt{2}}dx$$
so $w(x)=(1+2x)/\sqrt{2}$.
Remark: The above method only works for "nice" functions like $x^2$, for the width of the projection is sensitive to the slope. A function $f(x)$ for which the volume of the solid of revolution makes sense could have very unpleasant derivative, indeed $f(x)$ could fail to be differentiable.
To add a little detail, the point $(u,u^2)$ on the parabola projects to the point $((u+u^2)/2, (u+u^2)/2)$ on the line $y=x$.  Write down the results for $u=x$, and for $u=x+dx$, to find the coordinates of $P$ and $Q$, and use the Pythagorean Theorem to find the distance between $P$ and $Q$.
A: Too long for a comment:
The method mentioned by J.M. in the comments does indeed work, using a parametric representation of the function $y = x^2$ rotated counter-clockwise by $\pi/4$ radians. (The rotated parabola cannot be written as a function because rotating a parabola by an angle of $\theta$ is the same as rotating the axis of the parabola by $-\theta$, in a rotated frame of reference, which intersects the parabola at two points except when $\theta = 0, \pi$.)
The parametric form of $y = x^2$ is just $(t^2, t)$ for $t \in \mathbb R$. Now using the rotation matrix and taking advantage of the fact that $\cos \pi/4 = \sin \pi/4 = \frac{\sqrt2}{2}$, we have $(t,t^2) \mapsto \frac{\sqrt2}{2} \left(t + t^2, t^2 - t \right)$.
The bounds of the volume of revolution are $0$ and $1$ as those are when $t^2 = t$ ($x$ being $0, \sqrt{2}$ is irrelevant). Lastly, we need to find $dx = \frac{d}{dx} (x) \ dx = \frac{d}{dt} \left(\frac{\sqrt2}{2} (t + t^2) \right) \ dt = \frac{\sqrt{2}}{2}(1 + 2t) \ dt$.
And with that, we are all set. Putting the pieces into place, we have exactly the same integral as in André Nicholas's answer. For completeness, this evaluates to:
$$\int_{\text{lower}}^{\text{upper}} \pi y^2 \ dx = \int_0^1 \pi \left(\frac{\sqrt2}{2} \left(t^2 - t \right) \right)^2 \cdot \frac{\sqrt2}{2}(1 + 2t) \ dt$$
$$= \int_0^1 \frac{\pi \sqrt2}{4} (t^4 - 2t^3 + t^2)(1 + 2t) \ dt$$
$$= \frac{\pi \sqrt2}{4}  \int_0^1 2t^5-4t^4+2t^3+t^4 - 2t^3 + t^2 \ dt$$
$$= \frac{\pi \sqrt2}{4} \left(\frac{2}{6} - \frac{3}{5} + \frac{1}{3} \right) = \boxed{\frac{\pi \sqrt2}{60}}$$
