The Volume of Intersecting Cylinders

Now for another classic, although not as "kewel" as the spiral example. This time, it was I who asked for the help: once upon a time in a calculus class, we were asked to compute the volume of the solid that is the intersection of two cylinders at right angles. This is a surprisingly subtle calculation, and one of the more difficult-to-visualize things. Fortunately, I've managed to reproduce it above, with some transparency so the intersection is a little bit more visible. Now, here is the full intersection itself without the extra cylinder parts (called the Steinmetz Solid):

It is said that the Chinese mathematician 祖沖之 (Zu Chongzhi) computed the volume of this solid in the 400s ... I don't know how he did it without calculus, which is the way we're going to do it today. To do this, we write the equations of the two cylinders, say of radius $R$, which we will assume to have their axes in the $xy$-plane:

\[ x^2 + z^2 = R^2 \] \[ y^2 + z^2 = R^2. \]
Let's see what the actual intersection is. We set the two equations equal to each other:
\[
x^2 + z^2 = y^2 +z^2
\]
which means $x^2 = y^2$. This in turn means $y = \pm x$. In the $xy$-plane, this looks like two intersecting lines in an "X" shape, which you can see by looking at the solid above from the very tippy top. Of course, this is not exactly what the intersection actually looks like: we still have that there is a $z$-coordinate, given by $z = \pm \sqrt{R^2-x^2} =\pm \sqrt{R^2-y^2}$, so that the complete parametrization of the curves (four of them corresponding to each choice of $+$ or $-$) are given by, for example
\[
x = t
\]\[
y = \pm t
\]\[
z = \pm\sqrt{R^2-t^2}
\]
for $t$ varying from $-R$ to $R$ (the maximum allowable range given that it has to lie on a cylinder). This gives the "corners" of the solid given above. The solid itself is given by stacking squares with edges running from one curve to another (you can also see that in the solid above). To calculate the volume, we integrate the area of cross-sections perpendicular to the $z$-axis:
\[
V=\int_{-R}^R A(z)\,dz = \int_{-R}^R s^2\,dz.
\]
Therefore, we must find the side length $s$, which requires us to solve things in terms of $z$: $z =\pm\sqrt{R^2-x^2}$ means $z^2 = R^2 - x^2$, or $x^2 = R^2 - z^2$, so finally $x = \pm\sqrt{R^2 - z^2}$. Since the sides of the square are the same on either side, we can choose the positive square root. Then, since one side of the square goes between the two different values of $y$ for a given $x$, we find $s = \sqrt{R^2 - z^2}-(-\sqrt{R^2 - z^2}) = 2\sqrt{R^2 - z^2}$. Finally, we can find the integral:
\[
V = \int_{-R}^R s^2\, dz = \int_{-R}^{R} 4(R^2-z^2)\,dz = \left.4R^2 z - \frac{4}{3}z^3\right|_{-R}^R = \left(8-\frac{8}{3}\right)R^3 = \frac{16}{3}R^3.
\]

Yitz..'s Kewel Spiral

Every once in a while, I get an "assignment" to calculate some interesting parametrization (because parametrizations are awesome like that). For example, the bowtie pasta was originally a request from a student. Today's goes way back, to someone asking a question on a forum in math way back when. He needed a spiral that started out and twisted its way through space as it got larger. In essence, a spiral that doesn't stay in the same plane. Obviously, polar curves are out of the question, since they are, well, in the plane. But what's one dimension up from polar coordinates? Spherical coordinates $(r,\varphi,\theta)$, of course (or if you're a physicist, $(r,\theta,\phi)$).¹ We fall back on a good ol' tool, parametrization.

To discover what its equation might be, we first consider an Archimedean spiral ($r = a\varphi$) in the $r\varphi$-plane (or $yz$-plane), or possibly a logarithmic one, like $e^{\varphi/3}$, which grows as it goes around (like some spirals last time) the "pole" which is now the $x$-axis. This ends up growing as you move clockwise, rather than counterclockwise, since $\varphi$ increases as it goes from north toward east.

This is most efficiently represented in a parametric manner: first, in the $yz$-plane:\[
x=0,
\] \[
y = at \sin t,
\] \[
z = at \cos t.
\]
gives the parametric equation of the Archimedean spiral in the $yz$-plane going clockwise. This gives $\varphi = t$ and $r = at$, which recovers the original polar equation $r = a\varphi$.

Archimedean Spiral in the $yz$-plane, with $a = 1/20$.

How do we give it a twist? We reason that, after each revolution, we want it twisted out of the plane by some small amount (say, rotated about the $z$-axis by $b$ times a full revolution): we apply the rotation to the $x$ and $y$, giving:\[
x=at\sin t \cos (bt),
\] \[
y = at \sin t \sin(bt),
\] \[
z = at \cos t.
\]

Twisting out of the plane with parameter $b = 1/20$, also.

Here's the original one I delivered to Yitz.. (it is actually exponential):

Pretty kewel, right? (that was his exact word to describe it).

¹ I always use the convention that if I indeed need to refer to a physics text, do calculations to be consistent... but I always write my $\phi$'s as non-curly in that context... I haven't yet figured how to draw my $\theta$s different though. Yes, I know there is $\vartheta$, but it hasn't caught on in my notation. It just doesn't feel enough like a $\theta$.