The volume of this solid is the squared $\mathscr{L}^2$ distance between $y=x^2$ and $y=x^3$ on the interval $[0,\frac{3}{2}]$. |

The good ol' $\mathscr{L}^2$ norm has an outsized importance in many applications of math, principally, because Hilbert spaces (spaces with inner product) are useful for, well, just about anything. Much of a fan of Hilbert spaces I might have been for much of my recent life, it is only when teaching how to find the volume of solids (and a little help from Math Homie William) when it hit me what the best visualization for them is, at least for functions defined on $\mathbb{R}$. (For the longest time, I just said "It's sort of like the area between two curves, but, you know, squared, and stuff.") But we learn in integration theory that we can base a solid by putting squares (or semidisks or triangles, or whatever) in a perpendicular, 3rd dimension, over the area between curves; the volume of the solid is \[

V=\int_a^b A(x) \, dx

\] where $A(x)$ is the area of the cross section. But of course there's an obvious candidate: if you choose

*squares*to have side length being the height of your function $f(x)$, then $A(x)$ is just $f(x)^2$. Thus,\[

V = \int_a^b f(x)^2 dx.

\] But that's none other than the squared $\mathscr{L}^2$ norm of $f$! Similarly, for the volume of a solid consisting of squares between two curves $f(x)$ and $g(x)$, this is just \[

V=\int_a^b (f(x) - g(x))^2dx,

\] the (squared) $\mathscr{L}^2$ distance between the two functions. Now if we treat each of these squares like a deck of cards, and rotate them, it shouldn't change the volume. So we apply this to the functions $f(x)=x^2$ and $g(x)=x^3$ on $[0,\frac{3}{2}]$:

Now making vertical segments between the curve, and making a square with that side length pop half out of the plane, and half into the plane, gives us a nice solid. Rotating it gives the picture at the top.

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