If I take a one-to-one function $f(x)$ and rotate it about the x-axis, how can I describe a function resulting from a cross-section of the solid of revolution? I'm not talking about the circular cross-section; I'm talking about the other one. In other words, if I look at a plane intersecting the solid from above the Cartesian plane by a distance $d$, how can I describe the higher (bigger $y$) (the lower is simply minus the higher) intersection between the plane and the solid?
I'm wondering about this because I thought about the 2 well known methods for finding the volume of a solid of revolution: the cylindrical shell method and the washer/disk method. I wanted to come up with another method that's different, and I don't care that this method I came up with is way harder. Basically, I'm thinking of choosing any random cross-section and finding the area on that plane (needs an integral), and taking the integral of that over the entire solid.
If we were to take a plane intersecting at a distance $d$ from the Cartesian plane, and look at this from an angle, we'd get:
We're looking for an equation for $y^*$, which can be found by the Pythagorean Theorem: ${y^*}^2 + d^2 = f(x)^2$ so $y^* = \sqrt{f(x)^2-d^2}$.
Now the volume of your solid of revolution is:
$$ \int_{-max(f(x))}^{max(f(x))}2\int_{A(d)}^{B(d)}\sqrt{f(x)^2-d^2}\,\mathrm dx\,\mathrm dd $$
Where $max(f(x))$ is the maximum value of $f$ on the range of $x$ values, and $A(d)$ and $B(d)$ are the places where $y^* = 0$ or cuts off (there won't be more than these, because $f(x)$ is one to one). Obviously, the bounds will need tweaking for the various functions, but that integral is the general picture of the volume integral you want.