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!
 A: 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 $$V = \int_{x=a}^b dV = 2\pi h \int_{x=a}^b x - \frac{x^2}{b} \, dx = \frac{(a-b)^2(2a+b)h \pi}{3b}.$$
