Volume By Slicing & Revolutions - Generalizations? In single variable calculus there are the techniques of cross-sectional areas, volume by slicing, & volumes of solids of revolution (disks, washers, shells). Are these special cases of some concepts in double &/or triple intergals or are they just convenient tricks one can do with single variable integrals? A reference would especially be appreciated if possible - thanks!

These are all examples of computing a volume integral by breaking it down into lower-order integrals using a foliation of the region by surfaces. When these surfaces and the function you're integrating have appropriate symmetries, the double integrals are easily computed, leaving you with the single-integral formulae. It's easiest if we describe such a foliation by the level sets of a function $r$ - some examples:

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*$r(x,y,z) = \sqrt{x^2 + y^2}$ gives the cylinders around the $z$ axis for the "shell method";

*$r(x,y,z) = z$ gives the horizontal slices;

*$r(x,y,z) = \sqrt{x^2 + y^2 + z^2}$ gives the concentric spheres.

In its most general form, this technique is enabled by the so-called Co-area Formula (which in the smooth case comes from the change of variables formula and Fubini's theorem):

Let $\Omega \subset \mathbb{R}^n$ with a Lipshitz function $r:\Omega \to \mathbb{R}$, and let $\Sigma_t$ be the level set $\{ x \in \Omega : r(x) = t \}$. Then for any integrable function $f$ on $\Omega$ we have $$\int_\Omega f |\nabla r| dV = \int_{-\infty}^\infty  \int_{\Sigma_t} f dS  dt.$$
Here $dV$ is the Lebesgue (volume) measure and $dS$ is the $n-1$ dimensional Hausdorff measure (surface area in the case $n=3$) measure.

When the foliation is smooth ($|\nabla r| \ne 0$) and we take $f=1$, this gives the formula for volume:
$$ \int_\Omega dV = \int_{-\infty}^\infty \int_{\Sigma_t} \frac{1}{|\nabla r|}dS dt,$$
If you've chosen a nice foliation with $|\nabla r|$ constant on each level set, then this reduces to a single integral of an expression involving the area of the level sets. The three examples I gave above in fact have $|\nabla r|=1$ everywhere; so you get the easy formula $\mathrm{Vol(\Omega)} = \int \mathrm{Area}(\Sigma_t) dt$.
