# Dealing with the volume of a cone with a cylindrical hole bored in it

A cylindrical hole of radius $a$ is bored through a solid right-circular cone of height $h$ and base radius $b>a$. If the axis of the hole lies along that of the cone, find the volume of the remaining part of the cone.

Okay. So obviously I need to find the volume of the cone and then subtract the cylindrical hole. I'm having trouble visualizing the problem, and got caught up in the language used. I picture the cone as lying around the x-axis with a cyclinder hole right down the middle of it.

Volume of the whole cone is $V=\frac13\pi hb^2$ so if I can set up the circle, I can use $a$ as the radius and rotate it about $y=0$

Right now I've got $\frac\pi3b^2h-\pi a^2h=\frac\pi3h[b^2-3a^2]$
Do I integrate this?

I'm looking for someone to help set this up, even partially. Thank you!

## 1 Answer

Here's a hint. The volume of the portion of the cone that was bored out is equal to the sum of the volumes of two subregions: the first subregion is a cylinder, and the second is a smaller cone that sits atop it with the same radius. The small cone is similar in proportion to the large cone.

So, if the hole is of radius $a$, the big cone has height $h$ and base $b$, then the small cone has height $$h^* = \frac{a}{b}h$$ and radius $a$. The cylindrical portion of the hole has radius $a$, but what is its height? This is a little trickier. The key is to see that the height of the small cone plus the height of the cylinder must equal the total height of the big cone, $h$. That is to say, the cylinder's height must be $h - h^*$. Now that you can calculate the total volume the bored out solid, you can subtract it from the total volume of the cone.

Incidentally, a solution via the integration of cylindrical shells is as follows. Place the big cone on the coordinate axes, so that the $y$-axis coincides with the axis of the cone. Then consider the line that passes through the points $(b,0)$ and $(0,h)$, the latter point being the apex of the cone. This line has equation $$y - 0 = \frac{h-0}{0-b}(x - b) = -\frac{h}{b}x + h = h(1 - x/b).$$ Now consider the rotation of the triangular region enclosed by $$0 \le y \le h(1-x/b), \quad 0 \le a \le x \le b$$ about the $y$-axis. For a representative cylindrical shell with radius $x$ and height $f(x) = h(1-x/b)$, the differential volume of this shell is $$dV = 2 \pi x f(x) \, dx = 2 \pi h x(1-x/b) \, dx.$$ Therefore, the total volume of the cone with the cylindrical hole bored out is $$V = \int_{x=a}^b dV = 2\pi h \int_{x=a}^b x - \frac{x^2}{b} \, dx = \frac{(a-b)^2(2a+b)h \pi}{3b}.$$

• Makes sense, I was concerned about that. See my edit as to where I got. Did I simply not account for the "mini-cone"? – Yeah.. Mar 1 '16 at 1:33
• I've got it now. My only hang-up is how you got $h^*=\frac{a}{b} h$ – Yeah.. Mar 1 '16 at 1:42
• @Yeah.. Because the small cone is proportional to the large cone, since their bases are parallel disks and they share the same axis. Corresponding parts of similar objects are in proportion to each other; therefore, $h^*/h = a/b$. The height of the small cone divided by the height of the big cone equals the radius of the small cone divided by the radius of the big cone. – heropup Mar 1 '16 at 1:47