A New Angle On Intersecting Cylinders

This is a problem where the math is slice-integration with a very simple integral, but the visual thinking needed to figure out the shape of the slices is a good challenge.

Let the origin be where the axes of the cylinders cross, let the $z$ axis be the line perpendicular to both cylinders, and let the $y$ axis be the axis of one cylinder (the axis of the other is also in the $x$ - $y$ plane). We can use the equation of a circle or just the Pythagorean Theorem to determine that the points on the surface of that cylinder satisfy this equation:

$x = \sqrt{1-z^2}$

We will slice up the intersection of the cylinders in slices that are parallel to the x-y plane (the plane that contains the axes of both cylinders) and so have constant $z$ values, and we will integrate over $z$ from $-1$ to $1$ .

Consider first the $y$ -axis cylinder. At each $z$ , a slice of that entire cylinder is a rectangle running the length of the cylinder. That rectangle is $z$ from the origin, and its long edges are lines on the surface of the cylinder parallel to the $y$ axis and with constant $x$ values $\pm\sqrt{1-z^2}$ , and so the rectangle has width $2\sqrt{1-z^2}$ . A slice of the other entire cylinder at $z$ is of course exactly the same thing: a rectangle of the same width, in the same plane, except at 45 degrees relative to the first rectangle. The intersection of these two long rectangles is our desired slice, at $z$ , of the intersection of the cylinders. It is a rhombus with acute angle of 45 degrees. The rhombus has height $2\sqrt{1-z^2}$ and base $2\sqrt{2}\sqrt{1-z^2}$ , and its area is its base times its height.

So our answer is:

$\int_{-1}^{1} 4\sqrt{2}(1-z^2) dz = 4\sqrt{2} \bigg[_{-1}^1 z - \frac{z^3}{3} \bigg] =\sqrt{2} \frac{16}{3} \approx \boxed{7.54}$

By the way, this is $\sqrt{2}$ times the volume of the intersection of perpendicular cylinders. In general, the multiple is the secant of the angle of intersection.