Determine the Volume of Gabriel's horn with a lodged sphere Just recently I was posed with the following interesting question:
Say you took a sphere with an arbitrary, constant radius "$R$" and dropped it into Gabriel's Horn. Then you filled the horn perfectly with water such that the sphere acts like a plug and stops water from passing. What is the volume of the water that was poured into the horn?
The primary trouble is determining where it is that the sphere gets lodged in the horn, and determining what percentage of the sphere is above the point at which it is lodged. I thought it would be interesting to see what sort of solutions there were.
 A: Let's consider a cross-sectional case in 2 dimensions to see how deep it gets. Given the symmetry of both cross-sections about the $x$-axis, we really only need to deal with the portion above the axis. 
We'll be dealing, then with a curve of form $$y=\sqrt{R^2-(x-x_0)^2},$$ and the curve $$y=\frac1x,\:x\geq1.$$ The $x_0$ will be chosen so that the two curves intersect in precisely one point. Note that if $R\geq\sqrt 2$, then this intersection will necessarily occur at the point $(1,1)$, in which case there will be no room for water in the horn at all. (Why?)
Let us now suppose that $0<R<\sqrt 2$. The two arcs will then intersect in one point if and only if they share a tangent line at that point. If we call that point $(x_1,y_1)$ then to find the volume of the water, we will need to rotate the following region about the $x$-axis, amd calculate the volume of revolution (washer method recommended): bounded above by $y=\frac1x$, below by $y=\sqrt{R^2-(x-x_0)^2}$ (or $y=0$ where this isn't defined), on the left by $x=1$, and on the right by $x=x_1$. 
