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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?

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One thing you could do is to rotate your coordinate system such that the once oblique lines are parallel to the coordinate axes. It helps if you first express the curve you're rotating about the axis in parametric form... – J. M. Feb 2 '12 at 3:29
But that's avoiding the problem... – jak Feb 2 '12 at 3:48
There's really no point in persisting in one coordinate system when there's another that makes manipulations vastly easier. – J. M. Feb 2 '12 at 4:02

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 $Q$.

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How could u = x and u = x + dx at the same time? – jak Feb 2 '12 at 4:52
@jak: It isn't. In the formula for the projection of a point on the parabola onto the line $y=x$, we want to find two points, that I called $P$ and $Q$. Here $P$ is the projection of $(x,x^2)$ and $Q$ is the projection of $(x+dx,(x+dx)^2)$. So we need to apply the formula for two different values of $u$. I hope this clarifies things. It answers your "slanted differential" question. – André Nicolas Feb 2 '12 at 4:59

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