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 $Q$.