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.

\]

## No comments:

## Post a Comment